Exemplar MCQs

Hint

A bangle is in the form of a ring. The centre of mass lies at the centre , which is outside the body (boundary).

Hint

The position of centre of mass of the system shown in Fig. is likely to be at C. This is because lower part of the sphere containing sand is heavier than the upper part of the sphere containing air. 

Hint

The initial velocity is
\(
\vec{v}_i=v \hat{e} y
\)
After reflection from the wall, the final velocity is
\(
\vec{v}_f=-v \hat{e} y
\)
The trajectory is given as
\(
\vec{r}=y \hat{e}_y+a \hat{e}_z
\)
Hence, the change in angular momentum is
\(
\begin{aligned}
&\Delta \vec{L}=\vec{r} \times m\left(\vec{v}_f-\vec{v}_i\right) \\
&=\left(y \hat{e}_y+a \vec{e}_z\right) \times(-2 m \hat{e} y) \\
&=2 \operatorname{mva} \vec{e}_x\left[\therefore \hat{e}_y \times \hat{e}_y=0 \text { and } \hat{e}_z \times \hat{e}_y=-\hat{e}_x\right]
\end{aligned}
\)

Hint

When a disc rotates with uniform angular velocity, the angular acceleration of the disc is zero. Choice (d) is not true.

Hint

According to the theorem of perpendicular axes, \(I_z=I_x+I_y\). With the hole, \(I_x\) and \(I_y\) both decrease. Gluing the removed piece at the centre of the square plate does not affect \(I_z\). Hence, \(I_z\) decreases, overall.

Hint

As the piece is removed from quadrant I, therefore the center of mass of the square is shifted to the opposite of quadrant I i.e., quadrant III.

Hint

The density of non-uniform rod is given by
\(
\rho(x)=a\left(1+b x^2\right)
\)
So if \(b=0\), the density of the rod would be uniform and its centre of mass would lie at the middle point of the length of the rod, i.e.) \(0.5\)
Putting \(b=0\) for all the options, we get \(0.5\) only for \(
\frac{3(2+b)}{4(3+b)}
\)

Hint

Aa no external torque acts on the system, angular moemtum should be conserved.
Hence, \(I \omega=\) constant ….. (i)
Where , \(I\) is moment of inertia of the system and \(\omega\) is angular velocity of the system
From eq. (i) \(I_1 \omega_1=I_2 \omega_2\)
(where \(\omega_1\) and \(\omega_2\) are angular velocities before and after jumping)
\(
\Rightarrow I \omega=\frac{I}{2} \times \omega_2
\)
(as mass reduce to half, hence, the moment of inertia is also reduced to half)
\(
\omega_2=2 \omega
\)

Hint

(a) For a general rotational motion, angular momentum \(L\) and angular velocity \(\omega\) need not be parallel.
(c) For a general translational motion, momentum \(p\) and velocity \(v\) are always parallel.

Hint

The correct answer is the option (a) and (d)
Using \(\vec{L}=\vec{r} \times \vec{p}\) the direction of \(\mathrm{L}\) is determined as perpendicular to the plane of \(\mathrm{r}\) and \(\mathrm{p}\) by Right-hand thumb rule.
\(\overrightarrow{L_1}=\vec{r} \times m \vec{v}\) (out of the page) \(=m \vec{v} d_1\)
\(\overrightarrow{L_2}=\overrightarrow{r_2} \times m \overrightarrow{(-v)}\) (into the page) \(=m \vec{v} d_2\)
Hence \(\mathrm{a}\) is correct and b wrong.
For total angular momentum \(\vec{L}=\overrightarrow{L_1}+\overrightarrow{L_2}\)
\(\vec{L}=m \vec{v} d_1\) (out of page) \(-m \vec{v} d_2\) (into page)
\(\vec{L}=m \vec{v}\left(d_2-d_1\right)\) [ into the page]

Hint

When net external torque on a system of particles about an axis is zero, i.e.., \(\vec{\tau}=\vec{r} \times \vec{F}=r F \sin \theta \hat{\tau}=\) Zero, where, \(\theta\) is angle between \(\vec{r}\) and \(\vec{F}, \hat{\tau}\) is unit vector along \(\vec{\tau}\), them all the four statements (a),(b),(c),(d) are compatible.

Hint

By right-hand thumb rule, in the formula \(\vec{\tau}=\vec{r} \times \vec{F}\) the direction of \(\tau\) is perpendicular to the plane of \(\vec{r}\) and \(\vec{F}\)
\(\overrightarrow{\tau_z}=\vec{r} \times \vec{F}=r F \sin \theta \hat{k}\)
So a is incorrect
\(\overrightarrow{\tau_{z^{\prime}}}=\overrightarrow{r^{\prime}} \times \overrightarrow{F^{\prime}}=-r F^{\prime} \sin \theta \hat{k}\)
So \(b\) is correct
(c) Torque \(\tau\) caused by \(F\) about \(z\) axis is greater in magnitude than that about \(z\) axis.
\(
\begin{aligned}
&\because \mathbf{\tau}>\tau^{\prime} \\
&\theta>\theta^{\prime} \\
&\sin \theta>\sin \theta^{\prime} \\
&\tau F \sin \theta>\tau^{\prime} F \sin \theta \\
&\tau_z>\tau_x
\end{aligned}
\), hence c is correct.

\(\tau=\tau_z+\tau_{z^{\prime}}\) is only valid if \(\tau_z\) and \(\tau_{z^{\prime}}\) are along the same axis, which is not true in this case. Hence d is incorrect.

Hint

The correct answer is the option (a), (b), and (d).
By perpendicular axis theorem \(I_z=I_x+I_y\)
So a is correct
\(Z\) and \(z^{\prime}\) axis are parallel, and the distance between them can be calculated as \(\frac{D G}{2}\)
\(\frac{D G}{2}=\frac{1}{2} \sqrt{a^2+a^2}=\frac{a}{\sqrt{2}}\)
By parallel axis theorem \(I_z^{\prime}=I_z+M\left(\frac{a}{\sqrt{2}}\right)^2=I_z+\frac{M a^2}{2}\)
So \(b\) is correct
Since Z-axis and z” axis are not parallel to each other, so parallel theorem is not applicable here, rendering option c wrong
Since \(Z\) axis passes through the centre of the cube, so \(x\) and y-axis are symmetric \(I_x=I_y\)
Hence, \(d\) is correct

Hint

When the vertical height of the object is very small as compared to the earth’s radius, we call the object small, otherwise, it is extended. Building and pond are small objects. A deep lake and a mountain are examples of extended objects.

Hint

As, moment of Inertla of a given body \(\mathrm{I}=\sum m_i r_i^2\) (sum of M.I. of each constituent particles)

All mass in a cylinder lies at a distance R from the axis of symmetry but most of the mass of a solid sphere lies at a smaller distance than \(\mathrm{R}\).

That’s why a solid sphere has a smaller moment of Inertia than a hollow cylinder.

Hint

\(\omega=\frac{d \theta}{d t}\)
Here the \(\theta-t\) graph has a positive slope, so \(\omega\) is also positive,  which by convention, represents anticlockwise rotation.

Hint

From figure, Moment of force \(F\) about \(A\),
\(\tau_1=F \times a\), anticlockwise.
Moment of weight \(M g\) of cube about \(A_1\)
\(\tau_2=M g \times \frac{a}{2}\), clockwise.
The cube will not exhibit any motion,
if \(\tau_1=\tau_2\)
or \(F \times a=M g \times \frac{a}{2}\) or \(F=\frac{M g}{2}\)
The cube will rotate only, when \(\tau_1>\tau_2\)
\(F \times a>M g \frac{a}{2}\) or \(F>\frac{M g}{2}\)
If we assume that normal reaction is effective at \(a / 3\) from \(A\), then block would turn if
\(M g \times \frac{a}{3}=F \times a\) or \(F=\frac{M g}{3}\).
When \(F=\frac{M g}{4}<\frac{M g}{3}\), there will be no motion.

Hint

The sphere rolls without slipping when \(\omega=\frac{v}{r}\)
Let the velocity of the sphere after applying \(F\) be \(V\)
Then by the law of conservation of angular momentum
\(
\begin{aligned}
&m v(h-R)=I \omega \\
&m v(h-R)=\frac{2}{5} m R^2 \frac{v}{R} \\
&h-R=\frac{2}{5} R \\
&h=\frac{2}{5} R+R=\frac{7}{5} R
\end{aligned}
\)
Therefore, the sphere rolls without slipping with a constant velocity and no loss of energy. Thus (d) -(i)
Torque due to force \(F=\tau=(h-R) \times F\)
If \(\tau=0 \cdot h-R=0\) and thus \(h=R\)
In this case, the sphere will only have a translation motion and slip against the force of friction. Thus (b)-(iv)
For clockwise rotation of sphere \(\tau>0\)
\(
(h-R) \times F>0
\)
Or \(h>R\). Thus (c) – (ii)
For anti-clockwise rotation \(\tau<0\)
\(
(h-R) \times F<0
\)
\(h<R\), Thus (a) – (iii)

Hint

Given :
\(
\sum_i \vec{F}_i \neq 0
\)
Sum of torque about a certain point A,
\(
\sum_i \vec{r}_i \times \vec{F}_i=0
\)
Sum of torques about any other point B,
\(
\sum_i\left(r_i \overrightarrow{-} a\right) \times \vec{F}_i=\sum_{i=1}^n \vec{r}_i \times \vec{F}_i-a \sum_{i=1}^n \vec{F}_i
\)
So, as a and \(\sum_{i=1}^n \vec{F}_i\) are not zero.
\(\therefore\) torque is not necessarily zero about any other point.

Hint

No. A force can produce torque only along a direction normal to itself as \(\tau=r \times f\). So, when the door is in the xy-plane, the torque produced by gravity can only be along \(\pm\) z direction, never about an axis passing through y direction.

Hint

Hint: \(\sum_i m_i \overrightarrow{r_i}\) is zero at the center of mass.
Find the position vector of the center of mass By equating \(\sum_i m_i \overrightarrow{r_i}=0\).
Let \(\mathrm{x}\) be the distance of center of mass.
Now,
\(
\begin{aligned}
&(n-1) m x+m a=0 \\
&(n-1) m x=-m a \\
&x=- a/(n-1)
\end{aligned}
\)
The center of mass will get shifted in the opposite direction with reference to the position vector of the vacant vertex.

Hint

For the translational motion:
\(\mathrm{F}-\mathrm{f}=\mathrm{ma} \ldots \ldots \ldots . . .(1)\)
For the rotational motion:
\(
\begin{aligned}
f R &=I \alpha \\
f R &=\frac{m^2}{2} \times \frac{a}{R} \ldots \ldots .(2)
\end{aligned}
\)
Form equation 1 & 2
\(
\mathrm{f}=\frac{\mathrm{mg}}{2} \mathrm{f}=\mu \mathrm{mg}=\frac{\mathrm{mg}}{2}
\)
\(
\mathrm{ma}=2 \mu \mathrm{mg}
\)
\(
\mathrm{F}-\frac{\mathrm{mg}}{2}=\mathrm{mg}
\)
\(
\mathrm{F}=\frac{3 \mathrm{ma}}{2}=\frac{3 \times 2 \mu \mathrm{mg}}{2}
\)
\(
F=3 \mu \mathrm{mg}
\)

Hint

B implies A but A does not imply B because the linear momentum of the system remains constant if there is no net external force acting on the system. But center of mass of the system is at rest, its acceleration is zero and thus by newton’s second law of motion, the internal force is also zero implying the law of conservation of momentum. option -d is correct.

Hint

If \((B)\) is true, then \(1 / 2 \sum_i m_i v_i^2=0\). In this equation \(v\) is magnitude of velocity and \(m\) is mass. Mass cannot be zero and the square of a scaler quantity can only be zero if it is zero. It means the magnitude of the velocity of each particle is zero. In that case \(\Sigma_i m_i v_i=0\). So clearly (B) implies (A).

Now if (A) is true, it does not mean that magnitude of each particle is zero. Since linear momentum is a vector, and sum (resultant) of vectors can be zero even if each vector is non-zero. It means the momentum of each particle is not zero, hence some of the particles may have non-zero magnitude. In that case \((B)\) is not true. So (A) does not imply (B).

Hint

As the velocity of the particle depends on the frame of reference, the linear momentum as well as the kinetic energy is dependent on the frame of reference.

Hint

Distance of the centre of mass from the origin is given by,
\(
R^{\prime}=\frac{1}{M} \sum_{i=1}^n m_i \vec{r}_i
\)
Let half of the particles lie on \(+Y\)-axis and the rest of the particles lie on \(+X\)-axis.
\(
\begin{aligned}
Y_{\mathrm{CM}} &=\frac{m_3 R \vec{j}+m_4 R \vec{j}}{m_1+m_2+m_3+m_4+\ldots} \\
&=\frac{R \vec{j}}{2}
\end{aligned}
\)
Similarly, \(X_{\mathrm{CM}}=\frac{R \vec{i}}{2}\)
Therefore, the coordinates of centre of mass are \(\left(\frac{R}{2}, \frac{R}{2}\right)\)
Distance of the centre of mass from the origin \(=\frac{R}{\sqrt{2}}\).
For the given situation, \(R^{\prime}<R\).
In general, \(R^{\prime} \leq R\)

Hint

Area of circular plate, \(A_1=\pi\left(\frac{a}{2}\right)^2\)
\(x\) – coordinate of \(\operatorname{COM}, x_1=-\frac{a}{2}\)
Area of square plate, \(A_2=a^2\)
\(x\) – coordinate of \(\mathrm{COM}, \mathrm{x}_2=+\frac{\mathrm{a}}{2}\)
\(
\begin{aligned}
&\Rightarrow x_{C M}=\frac{A_1 x_1+A_2 x_2}{A_1+A_2} \\
&\Rightarrow x_{C M}=\frac{\pi\left(\frac{a^2}{2}\right)^2 \times\left(-\frac{a}{2}\right)+a^2\left(\frac{a}{2}\right)}{\pi\left(\frac{a^2}{2}\right)^2+a^2} \\
&\therefore x_{C M}=\frac{a(4-\pi)}{2(\pi+4)}
\end{aligned}
\)
It clearly shows that \(\mathrm{x}_{\mathrm{CM}}>0\), hence the COM of the composite system lies inside the square plate.

Hint

Acceleration of centre of mass of a two-particle system is given as,
\(
\vec{a}_{c m}=\frac{m_1 \vec{a}_1+m_2 \overrightarrow{a_2}}{m_1+m_2}
\)
According to the question,
\(
\begin{aligned}
&m_1=m_2=m \\
&a_1=0 \\
&a_2=a
\end{aligned}
\)
Substituting these values in equation (1), we get:
\(
\vec{a}_{c m}=\frac{m \times 0+m \vec{a}}{2 m}=\frac{1}{2} \vec{a}
\)

Hint

The answer is b. If there is no external force the linear momentum is conserved because the velocity of CoM is not changed, so internal forces cannot change the linear momentum of the system. Since internal forces can change the velocity of the constituent particles of the system, K.E. of the system can change because even for negative velocity K.E. \(=1 / 2 \mathrm{mv}^2\) will be positive.

Hint

When the block kept at rest is hit by a bullet, the block acquires certain velocity by the conservation of linear momentum. Therefore, the linear momentum and the kinetic energy of the block change.
As some of the kinetic energy carried by the bullet transforms into heat energy, its temperature also changes. However, the gravitational potential energy of the block does not change, as the height of the block does not change in this process. The answer is c.

Hint

we can use \(\mathrm{F}=\mathrm{ma}\), to know acceleration is dependent on force and mass and not on height So, from the formula we can find that the acceleration of the centre of mass of the sphere is independent of \(\mathrm{h}\), acceleration is independent of \(\mathrm{h}\)

Hint

The answer is b. As the body falls vertically downwards, no external force acts in the horizontal direction. Hence, the centre of mass does not shift horizontally.

Hint

The answer is b. No external force is acting on the (ball+box) system when the ball collides with the wall of the box i.e \(\vec{F}_{\text {ext }}=0\)
Thus acceleration of centre of mass of the system \(\overrightarrow{\mathrm{a}}_{\mathrm{cm}}=\frac{\overrightarrow{\mathrm{F}}_{\mathrm{ext}}}{\mathrm{M}}=0\) \(\Rightarrow \Delta \overrightarrow{\mathrm{v}}=0\). Thus the velocity of the (ball+box) system remains constant.

Hint

Let the velocities of two parts be \(v\) and \(v^{\prime}\) and mass of each part \(=m\). Momentum before the break \(=2 \mathrm{~m}{\times} 0=0\).

Momentum after the break \(=m v+m v^{\prime}\). Since there is no external force on the body, the linear momentum will be conserved.
\(
{mv}+{m} {v}^{\prime}=0
\)
\(
v=-{v}^{\prime}
\)
So the parts will move in opposite directions with equal speeds.

Hint

Since the heavy ring and the small particle have equal masses, the center of mass of the system will be at r/2 from the ring. Hence the force acting on the system \(=m v^2 /\left(r / 2)=2 m v^2 / r\right.\)

Hint

In a collision of two bodies, the momentum of the body is conserved, but kinetic energy is not conserved, as some part of the energy is wasted in form of friction energy, which increases the surface temperature of the body, thus temperature is also not conserved, hence in a collision of two bodies, the only physical property which is constant is momentum.

Hint

By the law of conservation of linear momentum, we can write: Initial momentum \(=\) Final momentum
\(\Rightarrow m \vec{v}=m_1 \vec{v}_1+m_2 \vec{v}_2\)
\(\Rightarrow\left(m_1 \vec{v}_1+m_2 \vec{v}_2\right)\) must be parallel to \(\vec{v}\)

Hint

Shell is fired with velocity \(v\) at an angle \(\theta\) with the horizontal.
So its velocity at the highest point \(=\) horizontal component of velocity \(=v \cos\theta\)
So momentum of shell before explosion \(=m v \cos \theta\)
When it breaks into two equal pieces and one piece retraces its path to the canon, let the other part move with velocity \(\mathrm{v_{1}}\).
So momentum of two pieces after explosion \(=\frac{m}{2}(-v \cos \theta)+\frac{m}{2} v_{1}\)
By the law of conservation of momentum \(m v \cos \theta=\frac{-m}{2} v \cos \theta+\frac{m}{2} v_{1} \Rightarrow v_{1}=3 v \cos \theta\)

Hint

The initial kinetic energy before the collision and the final kinetic energy after the collision will remain the same. We cannot say that (c) is true because during the contact a part of the kinetic energy is stored as elastic potential energy in the bodies, and at this time total kinetic energy is less than the initial or final kinetic energy.

Hint

If the collision is perfectly inelastic the deformation of the bodies during the collision is permanent and the part of the initial kinetic energy lost in deformation is not recovered. As a result, the final kinetic energy is less than the initial kinetic energy.  If the collision is partly inelastic, the deformations of the bodies are partly recovered, and still, a part of the initial kinetic energy remains utilized in the partial deformation. Hence here again the final kinetic energy is less than the initial kinetic energy. Though the difference, in this case, is less than the perfectly inelastic collision.

Hint

The answer is none.
The centre of mass of a system of particles depends on the product of individual masses and their distances from the origin.
Therefore, we may say about the given statements:
(a) Distance of particles from origin is not known.
(b) Masses are same but the distance of particles from the origin is not given.
(c) Distance of particles from the origin is not given.
(d) It is not necessary that at least one particle lies on the negative \(X\)-axis. The particles can be above the negative \(X\)-axis on \(X-Y\) plane.

Hint

(c) may be all non-negative
(d) may be positive for some cases and negative for other cases
According to the question, the centre of mass is at origin.
\(
\begin{aligned}
&\therefore X=\frac{m_1 x_1+m_2 x_2+m_3 x_3+\ldots}{m_1+m_2+m_3+\ldots \ldots}=0 \\
&\Rightarrow m_1 x_1+m_2 x_2+m_3 x_3+\ldots=0
\end{aligned}
\)
From the above equation, it can be concluded that all the \(x\)-coordinates may be non-negative. In other words, they may be positive for some cases and negative for others.

Hint

(a) the density continuously increases from left to right
(b) the density continuously decreases from left to right
As the density continuously increases/decreases from left to right, there will be difference in the masses of the rod that lie on either side of the centre of mass. Thus, the centre of mass of a rod in such a case will certainly not be at its centre.

Hint

If there is no/zero resultant force acting on a system, there will be no acceleration. Hence (b), not (d).

If it is already moving with a uniform velocity, it will continue its motion without acceleration. Hence (c), not (a).

Hint

(b) \(v_0=0, a_0 \neq 0\)
(d) \(v_0 \neq 0, a_0 \neq 0\)
If a non-zero external force acts on a system of particles, it causes the centre of mass of the system to accelerate with acceleration \(a_0[latex] at any instant [latex]t\). In such a case, the velocity of centre of mass of the system of particles is either \(v_0\) or zero.

Hint

Both the balls experience the gravitational force. This force is proportional to the mass of the ball.
Thus, acceleration is \(\frac{\text { Force }}{\text { mass }}\) is a constant and is equal to \(\mathrm{g}\).

Hint

(a) the total momentum must be conserved
(d) the total kinetic energy must change.
Explanation:
When the block moving in air breaks into two parts due to internal forces, the speeds of the parts will change. Since the Kinetic energy is proportional to the square of the speed it will be always positive. Hence the total kinetic energy must change.

The momentum is a vector and even if the velocities of the parts are different in magnitude and direction the vector addition of momentums will be same as before the split. Because in the absence of external force the total momentum is conserved.

Hint

(b) the linear momentum remains constant
(c) the final kinetic energy Is equal to the initial kinetic energy
(d) the final linear momentum Is equal to the Initial linear momentum.
Explanation:
In a fully elastic collision only during the contact period a part of the kinetic energy is stored as elastic potential energy in the bodies and their shapes are slightly changed. Since the collision is elastic the bodies release the stored elastic potential energy and regain their shape. Again the total energy is in the form of K.E. Hence the final K.E. is equal to the initial K.E. But we can not say that the K.E. remains constant because during the contact period it is reduced. So (c) is true and (a) false. When there is no external force the momentum is conserved. Hence (b) and (d) is true.

Hint

(c) the total momentum of the ball and the earth is conserved
(d) the total energy of the ball and the earth remains the same
As the ball rebounds after hitting the floor, its velocity changes.
i.e. Velocity of the ball before collision \(\neq\) Velocity of the ball after the collision
Therefore, the momentum of the ball just after the collision is not the same as that just before the collision. The mechanical energy of the ball also changes during the collision.

However, the total momentum of the system (earth plus ball) and the total energy of the system remain conserved.

Hint

(b) both the bodies move after collision
(c) the moving body comes to rest and the stationary body starts moving
Explanation:
Since one of the bodies has some velocity before collision hence total momentum of the bodies is nonzero. Since in the absence of external force total momentum is conserved, both the bodies cannot come to rest. Because then the total momentum will become zero. (a) is not true.

In both options (b) and (c) it is possible that the final momentum is equal to the initial momentum.

Option (d) is not possible because the total momentum will not remain the same as before the collision.

Hint

All of the above.
(a) the velocities are interchanged
(b) the speeds are interchanged
(c) the momenta are interchanged
(d) the faster body slows down and the slower body speeds up

Hint

The directions of \(\mathrm{\vec{A}}\) and \(\mathrm{\vec{B}}\) will be perpendicular to each other. Hence \(\mathrm{\vec{A}} \cdot \mathrm{\vec{B}}=\) \(|\mathrm{A}| \cdot|\mathrm{B}| \cdot \operatorname{Cos} 90^{\circ}=0\).

Hint

The unit vector along the axis of rotation and the unit vector along the resultant force on the particle are perpendicular to each other in a uniform rotation.
Therefore, we have
\(\vec{A} \cdot \vec{B}=|A| \cdot|B| \cdot \operatorname{Cos} 90^{\circ}=0\)

Hint

Consider the diagram.

As, the particle goes with constant linear velocity, the perpendicular distance of velocity vector from the axis is constant. So, angular momentum \(m v r \sin \theta\) is constant.

Hint

In a pure rotation, the angular velocity of all the particles remains the same and does not depend on the position of the particle from the axis of rotation.
\(v \propto r\) and \(\omega\) is the constant of proportionality. Thus \(v=\omega r\).
So \(\omega\) is independent of \(r\).

Alternate:
If \(r\) be the distance of point from axis will increase, then the corresponding \(v\) of the point will also increase accordingly, Hence, \(\omega\) is constant.

Hint

It is given that angular velocity is same for both the wheels.
Therefore, we have
\(
\begin{aligned}
& v_A=\omega R \\
& v_B=\omega 2 R
\end{aligned}
\)
\(
\mathrm{v}_{\mathrm{B}}=\omega 2 \mathrm{R}
\)
\(
x=v_A t=\omega R t \dots(1)
\)
\(
y=v_B t=\omega(2 R) t \dots(2)
\)
From equations (1) and (2), we get
\(
y=2 x
\)

Hint

Because the resultant force on a particle of the body rotating uniformly is always perpendicular to the rotation axis and passes through it.

Hint

The resultant force on a particle of the body rotating non-uniformly is always horizontal and skew with the rotation axis because the net torque on the body is non-zero.

Hint

We have
\(
\vec{\Gamma}=\vec{r} \times \vec{F}
\)
Thus, \(\vec{\Gamma}\) is perpendicular to \(\vec{r}\) and \(\vec{F}\). Therefore, we have
\(
\vec{r} \cdot \vec{\Gamma}=0 \text { and } \vec{F} \cdot \vec{\Gamma}=0
\)

Hint

Consider a small portion of rod at a distance \(x\) from the clamped end (as shown in fig.) with width \(\mathrm{dx}\) and mass \(\mathrm{dm}\).

Centripetal force on this portion \(=\omega^2 x d m\) \(d m=\left(\frac{m}{l}\right) l d x\)
Force on the whole rod \(=\mathrm{F}=\int_0^l \omega^2 x \frac{m}{l} d x\) \(\therefore F=\frac{1}{2} m \omega^2 l\)

Hint

It is given that there is no force along \(x\)-axis.
COM of rod will remain and will not shift along \(x\)-axis (horizontal direction).
Force gravity is acting along \(y\)-axis (vertical direction). So, COM will shift along the \(y\)-axis by \(\frac{l}{2}\) distance and COM of horizontal rod is at a distance \(\frac{l}{2}\) from one end. Therefore, lower end of the rod will remain at a distance \(\frac{l}{2}\) from O.

Hint

Moment of inertia of circular disc of radius \(r\) :
\(
I=\frac{1}{2} m r^2
\)
Mass \(=\) Volume \(\times\) Density
Volume of disc \(=\pi r^2 t\)
Here, \(t\) is the thickness of the disc.
As density is same for both the rods, we have
Moment of inertia,
\(
\begin{aligned}
& I \propto \text { thickness } \times(\text { radius })^4 \\
& \frac{I_A}{I_B}=\frac{t \cdot(r)^4}{\frac{t}{4}(4 r)^4}<1 \\
& \Rightarrow \frac{I_A}{I_B}<1 \\
& \Rightarrow I_A<I_B \\
&
\end{aligned}
\)

Hint

\(
\tau=I \alpha(\text { magnitude) }
\)
For equal torque, we have
\(
\begin{aligned}
& I_A \alpha_A=I_B \alpha_B \\
& \mathrm{I}_{\mathrm{A}}<\mathrm{I}_{\mathrm{B}} \\
& \Rightarrow \alpha_A>\alpha_B \dots(1)
\end{aligned}
\)
Now,
\(
\omega=\alpha t
\)
Or,
\(
\begin{aligned}
& \frac{v}{r}=\alpha t \\
& v_A>v_B \text { (Using (1)) }
\end{aligned}
\)

Hint

Moment of inertia of a mass is directly proportional to the square of the distance of mass from the axis of rotation. Therefore, we have
\(
I \propto r^2
\)
As the tube is rotated, water is collected at the end of tube because of centrifugal force and the distance from the rotation axis increases. Hence, the moment of inertia increases.

Hint

Consider an element of length, \(d l=r d \theta\).
\(
\begin{aligned}
& d m=\frac{M}{\pi r} d l=\frac{M}{\pi r} r d \theta \\
& \text { Moment Of Inertia of semicircular wire }=\int_0^\pi r^2 d m \\
& I=\int_0^\pi r^2 \frac{m}{\pi r} r d \theta \\
& \Rightarrow I=m r^2
\end{aligned}
\)

Hint

Let \(\mathrm{d}\) be the density of the material and \(\mathrm{V}, \mathrm{R}\) be the volume and radius respectively.
THen. the moment of inertia of the body is given as:
\(
\mathrm{I}=\mathrm{KMR}^2
\)
The mass of the body is also written as the product of its density and volume.
\(
\therefore \mathrm{I}=\mathrm{KdV} \mathrm{R}^2
\)
where \(\mathrm{K}\) is a constant number.
For same identical geometrical shape, \(\mathrm{I} \propto \mathrm{d}\)
Now as \(\mathrm{d}_{\text {iron }}>\mathrm{d}_{\text {aluminium }}\)
\(
\Rightarrow \mathrm{I}_1 \lt \mathrm{I}_2
\)

Hint

\(
I_{\mathrm{x}}=m_2 a^2+m_3 a^2=0.20 \dots(1)
\)
\(
I_{\mathrm{y}}=m_1 a^2+m_3 a^2=0.20 \dots(2)
\)
\(
I_z=m_1 a^2+m_2 a^2 \dots(3)
\)
We have three equations and four variables. So, \(\mathrm{I}_{\mathrm{z}}\) cannot be deduced with the given information.

Hint

Let \(\mathrm{N}\) be the normal reaction on the block.

From the free body diagram of the block, it is clear that the forces \(\mathrm{N}\) and mgcos \(\theta\) pass through the same line. Therefore, there will be no torque due to \(\mathrm{N}\) and \(\mathrm{mg} \cos \theta\). The only torque will be produced by \(\mathrm{mg} \sin \theta\).
\(
\therefore \vec{\tau}=\vec{F} \times \vec{r}
\)
Since \(a\) is the edge of the cube, \(r=\frac{a}{2}\)
Thus, we have
\(
\begin{aligned}
& \tau=m g \sin \theta \times \frac{a}{2} \\
& =\frac{1}{2} m g a \sin \theta
\end{aligned}
\)

Hint

No external torque is applied on the ring; therefore, the angular momentum will be conserved.
\(
\begin{aligned}
& I \omega=I^{\prime} \omega^{\prime} \\
& \Rightarrow \omega^{\prime}=\frac{I \omega}{I^{\prime}} \dots(1)
\end{aligned}
\)
\(
I=M r^2
\)
\(
I^{\prime}=M r^2+2 m r^2
\)
On putting these values in equation (1), we get
\(
\omega^{\prime}=\frac{\omega M}{M+2 m}
\)

Hint

\(
\text { The given solution can be shown as }
\)

Here \(v_0=R \omega\)
At \(P, v=r \omega\)
\(
\begin{aligned}
& =\sqrt{\left(R^2+R^2\right) \omega} \\
& =\sqrt{2} R \omega=\sqrt{2} v_0
\end{aligned}
\)

Hint

As the distance covered in one revolution about the centre is less than the perimeter of the wheel, it means that the direction of torque due to frictional force opposes the motion of wheel, i.e., the frictional force acting on the wheel by the surface is along the velocity of the wheel.

Hint

On a frictionless road, we have
Angular velocity of the engine =0.
Therefore, an increase in petrol input will not affect the angular velocity and hence the linear velocity of the scooter will remain the same.

Hint

The incline is smooth; therefore, all bodies will slip on the incline. Also, as the mass of bodies is same, they will reach the bottom in equal time.

Hint

Since linear acceleration is the same for all i.e.
\(
a=g \sin \theta-\mu g \cos \theta
\)
As they have the same mass \((m)\) and friction coefficient \((\mu)\) Distance covered, \(\mathrm{s}=\frac{1}{2} \mathrm{at}^2\)
Since all three starts from rest and cover equal distance while accelerating at same rate, we can conclude that all will reach the bottom simultaneously.

Hint

Torque is same for all the bodies; therefore, the angular momentum will be conserved.
Now, total kinetic energy \(=\frac{1}{2} m v^2+\frac{L^2}{2 I}\)
So, the body with greater value of moment of inertia will have smallest kinetic energy at the bottom of the incline.
Order of the moment of inertia of the bodies:
hollow sphere > disc > solid sphere
Hence, the hollow sphere will have the smallest kinetic energy at the bottom.

Hint

For pure rolling,
\(
\omega r=v_0
\)
As shown in the figure, the velocity of the string will be resultant of \(v_0\) and \(\omega r\).
\(
\begin{aligned}
& v_{n e t}=v_0+(\omega r) \\
& v_{n e t}=2 v_0 \\
\end{aligned}
\)
Let, Linear distance travelled by the cylinder in time \((t)\),
\(
v_0 t=l
\)
\(\therefore\) Linear distance travelled by the string in the same time,
\(
2 v_0 t=2 l
\)

Hint

(b) may pass through the centre of mass
(d) may pass through a particle of the body
It is not necessary that the axis of rotation of a purely rotating body should pass through the centre of mass or through a particle of the body. It can also lie outside the body.

Hint

\(A\) is true but \(B\) is false
In non-inertial frames,
Here, \(\Gamma_{\text {total }}\) is the total torque on the system due to all the external forces acting on the system. So, equation (B) is not true as in non-inertial frames, pseudo force must be applied to study the motion of the object.

Hint

(b) Angular momentum \(=m(\vec{r} \times \vec{v})\)
If the point is on the straight line, \(\vec{r}\) and \(\vec{v}\) will have the same direction and their cross product will be zero. Hence, angular momentum is zero.
(c) If the point is not on the straight line, \(\vec{r}\) and \(\vec{v}\) will not have the same direction and their cross-product will not be zero. Hence, angular momentum is non-zero.
(d) No external torque is applied on the body; therefore, its angular momentum about any given point remains constant.

Hint

(a) angular momentum
(b) linear momentum
\(
\begin{aligned}
& \vec{F}_{\text {ext }}=0 \\
& \Rightarrow \vec{\tau}_{\text {ext }}=0
\end{aligned}
\)
That is, the change in linear momentum and angular momentum is zero. This is because:
\(
\frac{d \vec{P}}{d t}=\vec{F}_{e x t}
\)
And \(\frac{\overrightarrow{d L}}{d t}=\vec{\tau}_{e x t}\)

Hint

If axes \(A\) and \(B\) are parallel, we get
\(
I_B=I_A+m r^2
\)
Here, \(r\) is the distance between two axes and \(m\) is the mass of the body.
\(
\therefore I_A \lt I_B
\)

Hint

The particles on the diameter mentioned above do not have any linear acceleration.
Explanation:
The linear acceleration of a rotating particle is given as
\(
\vec{a}=\vec{r} \times \vec{\alpha}
\)
The sphere is rotating about a diameter; therefore, the position vector of the particles on the diameter is zero. Thus, the linear acceleration of the particle is zero.

Explanation of other options: All the particles of the body have the same angular velocity. All the particle on the surface have different linear speeds that depend on the position of the particle from the axis of rotation.

Hint

The torque of the applied force does not depend on the density of a rod. It depends on the distance between the pivot and the point where \(F\) is applied. So, it does not depend on which end of the rod is pivoted.

Hint

For pure rolling, \(\omega r=v_0\)
The velocity of the particle at \(A, B, C\) and D will be the resultant of \(v_0\) and \(\omega r\).

At point \(B\),
\(
\begin{aligned}
& v_{\text {net }}=\sqrt{v_0^2+(\omega r)^2} \\
& v_{\text {net }}=\sqrt{v_0^2+v_0^2} \\
& v_{\text {net }}=\sqrt{2} v_0
\end{aligned}
\)
At point \(\mathrm{C}\),
\(
\begin{aligned}
& v_{\text {net }}=v_0+(\omega r) \\
& v_{\text {net }}=2 v_0 \\
& \text { At point A, }
\end{aligned}
\)
At point A,
\(
\begin{aligned}
& v_{\text {net }}=2 v_0 \\
& v_{\text {net }}=v_0-(\omega r)
\end{aligned}
\)
At point \(\mathrm{O}\),
\(
\begin{aligned}
& r=0 \\
& \therefore v_{\text {net }}=v_0
\end{aligned}
\)

Hint

Acceleration of a sphere on the incline plane is given by
\(
a=\frac{g \sin \theta}{1+\frac{I_{C O M}}{m r^2}}
\)
\(I_{C O M}\) for a solid sphere \(=\frac{2}{5} m r^2\)
\(S o, a=\frac{g \sin \theta}{1+\frac{2 m r^2}{5 m r^2}}=\frac{5}{7} g \sin \theta\)
\(a\) is independent of mass and radii; therefore, the two spheres reach the bottom together.

Hint

\(
a=\frac{g \sin \theta}{1+\frac{I_{C O M}}{m r^2}}
\)
\(I_{C O M}\) for a solid sphere \(=\frac{2}{5} m r^2\)
So, \(a=\frac{g \sin \theta}{1+\frac{2 m r^2}{5 m r^2}}=\frac{5}{7} g \sin \theta\)
\(I_{C O M}\) for a hollow sphere \(=\frac{2}{3} m r^2\)
So, \(a^{\prime}=\frac{g \sin \theta}{1+\frac{2 m r^2}{3 m r^2}}=\frac{3}{5} g \sin \theta\)
The acceleration of the solid sphere is greater; therefore, it will reach the bottom with greater speed.

Hint

(b) A smooth inclined surface.
A sphere cannot roll on a smooth inclined surface because there is no backward force (force of friction) to prevent its slipping.

Hint

Explanation:-
(a) On the rear wheels, friction force is in the forward direction because it favours the motion and accelerates the car in forward direction.
(b) Because of the movement of the car in forward direction, front wheels push the road in forward direction and in reaction, the road applies friction force in the backward direction.
(c) As the car is moving in forward direction, the rear wheels have larger magnitude of friction force (in forward direction) than on the front wheels.

Hint

The given coefficient of friction \(\left(\frac{1}{7} g \tan \theta\right)\) is less than the coefficient friction \(\left(\frac{2}{7} g \tan \theta\right)\) required for perfect rolling of the sphere on the inclined plane.
Therefore, sphere may slip while rolling and it will translate and rotate about the centre.

Hint

(a) decrease the linear velocity
(b) increase the angular velocity
If a sphere is rolled on a rough horizontal surface, the force of friction tries to oppose the linear motion and favours the angular motion.

Hint

Let’s look into the free-body diagram of the sphere.

Net force acting on the sphere along the inclined plane-
\(
F_{\text {net }}=m \times g \times \sin \theta-m \times a \times \cos \theta
\)
Now looking into the fig. we have
\(
\tan \theta=\frac{\mathrm{a}}{\mathrm{g}}
\)
\(
\Rightarrow \mathrm{a}=\mathrm{g} \times \tan \theta
\)
\(
\mathrm{F}_{\text {net }}=0
\)
So, the net force remains zero which means, if the sphere is set in pure rolling on the incline, it will continue pure rolling.