Rotational Motion — Page 10 | NEET Physics

Q91

If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the center of the Sun, its areal velocity is :

A

{L \over m}

B

{4L \over m}

C

{L \over 2m}

D

{2L \over m}

Correct Answer

Option C

Solution

dA =

{1 \over 2}

r2d $\theta$ $\therefore$

{{dA} \over {dt}} = {1 \over 2}{r^2}{{d\theta } \over {dt}}

$\Rightarrow$

{{dA} \over {dt}} = {1 \over 2}{r^2}\omega

. . . . . (1) We know, angular momentum, L =

mvr

=

m\left( {\omega r} \right)r

= mr2 $\omega$ $\therefore$ $\omega$ =

{L \over {m{r^2}}}

. . . . . (2) Put value of $\omega$ in equation(1),

{{dA} \over {dt}}

=

{1 \over 2}{r^2}

(

{L \over {m{r^2}}}

) =

{L \over {2m}}

Q92

A uniform solid cylinder of mass ' m ' and radius ' r ' rolls along an inclined rough plane of inclination

45^{\circ}

. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be

A

\sqrt{2} \mathrm{~g}

B

\dfrac{1}{\sqrt{2}} \mathrm{~g}

C

\dfrac{1}{3 \sqrt{2}} \mathrm{~g}

D

\dfrac{\sqrt{2} g}{3}

Correct Answer

Option D

Solution

The acceleration of the center of mass for a rolling object down an incline is given by:

a = \frac{g \sin\theta}{1 + \frac{I}{m r^2}}

For a uniform solid cylinder, the moment of inertia about its central axis is:

I = \frac{1}{2} m r^2

Substituting this into the expression for acceleration:

a = \frac{g \sin\theta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{g \sin\theta}{\frac{3}{2}} = \frac{2}{3} g \sin\theta

For an incline of

45^\circ

, we have:

\sin 45^\circ = \frac{\sqrt{2}}{2}

Thus, the acceleration becomes:

a = \frac{2}{3} g \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} g}{3}

Therefore, the linear acceleration of the cylinder's axis is:

\frac{\sqrt{2} g}{3}

Q93

Initial angular velocity of a circular disc of mass

M

is

{\omega _1}.

Then two small spheres of mass

m

are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?

A

\left( {{{M + m} \over M}} \right)\,\,{\omega _1}

B

\left( {{{M + m} \over m}} \right)\,\,{\omega _1}

C

\left( {{M \over {M + 4m}}} \right)\,\,{\omega _1}

D

\left( {{M \over {M + 2m}}} \right)\,\,{\omega _1}

Correct Answer

Option C

Solution

When two small spheres of mass

m

are attached gently, the external torque, about the axis of rotation, is zero. So,

{{d\overrightarrow L } \over {dt}} = \overline z

= 0

\overrightarrow L

= conserved So the angular momentum about the axis of rotation is conserved. $\therefore$

{I_1}{\omega _1} = {I_2}{\omega _2}

\Rightarrow {\omega _2} = {{{I_1}} \over {{I_2}}}{\omega _1}

Here Moment of inertia of Disc

{I_1} = {1 \over 2}M{R^2}

and After adding two sphere Moment of Inertia of disc and two sphere,

{I_2} = {1 \over 2}M{R^2} +

2\left( {{1 \over 2}m{R^2} + {1 \over 2}m{R^2}} \right)

$\therefore$

{\omega _2} = {{{1 \over 2}M{R^2}} \over {{1 \over 2}MR + 2m{R^2}}} \times {\omega _1} = {M \over {M + 4m}}{\omega _1}

Q94

Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter. The correct statement for this situation is

A All of them will have same velocity.

B The ring has greatest and the cylinder has the least velocity of the centre of mass at the bottom of the inclined plane.

C The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.

D The cylinder has the greatest and the sphere has the least velocity of the centre of mass at the bottom of the inclined plane.

Correct Answer

Option C

Solution

{{{K_T}} \over {{K_R}}} = {{M{R^2}} \over {{I_{CM}}}}

ICM is maximum for ring. $\Rightarrow$ v is least for ring.

Q95

A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that

\theta(t)=5 t^2-8 t

, where

\theta(t)

is the angular position of the rotating disc as a function of time

t

. How much power is delivered by the applied torque, when

t=2 \mathrm{~s}

?

A

60 \mathrm{MR}^2

B

72 \mathrm{MR}^2

C

8 \mathrm{MR}^2

D

108 \mathrm{MR}^2

Correct Answer

Option A

Solution

Moment of Inertia and Angular Motion For a solid circular disk, the moment of inertia is given by

I = \frac{1}{2}MR^2.

The angular position is defined as

\theta(t) = 5t^2 - 8t.

Angular Velocity and Acceleration Differentiate with respect to time to obtain the angular velocity:

\omega(t) = \frac{d\theta}{dt} = 10t - 8.

Differentiating again, the angular acceleration is:

\alpha(t) = \frac{d\omega}{dt} = 10.

At time

t = 2 \, \text{s}

: Angular velocity:

\omega(2) = 10(2) - 8 = 12 \, \text{rad/s}.

Angular acceleration:

\alpha(2) = 10 \, \text{rad/s}^2.

Torque Calculation The torque applied by the external force is related to the moment of inertia and angular acceleration:

\tau = I \alpha = \frac{1}{2}MR^2 \times 10 = 5MR^2.

Power Delivered Power delivered by a torque is given by:

P = \tau \omega.

At

t = 2 \, \text{s}

, substitute the values:

P = 5MR^2 \times 12 = 60MR^2.

Thus, the power delivered by the applied torque at

t = 2 \, \text{s}

is:

\boxed{60MR^2}.

Q96

One solid sphere

A

and another hollow sphere

B

are of same mass and same outer radii. Their moment of inertia about their diameters are respectively

{I_A}

and

{I_B}

such that

A

{I_A} < {I_B}

B

{I_A} > {I_B}

C

{I_A} = {I_B}

D

{{{I_A}} \over {{I_B}}} = {{{d_A}} \over {{d_B}}}

where

{d_A}

and

{d_B}

are their densities.

Correct Answer

Option A

Solution

For solid sphere the moment of inertia of

A

about its diameter

{I_A} = {2 \over 5}M{R^2}.

The moment of inertia of a hollow sphere

B

about its diameter

{I_B} = {2 \over 3}M{R^2}.

$\therefore$

{I_A} < {I_B}

Q97

A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is -

A

{F \over {2mR}}

B

{2F \over {3mR}}

C

{3F \over {2mR}}

D

{F \over {3mR}}

Correct Answer

Option B

Solution

FR =

{3 \over 2}

MR2 $\alpha$ $\alpha$ =

{{2F} \over {3MR}}

Q98

A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is :

A Solid sphere

B Solid cylinder

C Hollow cylinder

D Ring

Correct Answer

Option B

Solution

{1 \over 2}I{\omega ^2} = {1 \over 2} \times {1 \over 2}m{v^2}

I = {1 \over 2}m{R^2}

Body is solid cylinder

Q99

A round uniform body of radius

R,

mass

M

and moment of inertia

I

rolls down (without slipping) an inclined plane making an angle

\theta

with the horizontal. Then its acceleration is

A

{{g\,\sin \theta } \over {1 - M{R^2}/I}}

B

{{g\,\sin \theta } \over {1 + I/M{R^2}}}

C

{{g\,\sin \theta } \over {1 + M{R^2}/I}}

D

{{g\,\sin \theta } \over {1 - I/M{R^2}}}

Correct Answer

Option B

Solution

A uniform body of radius R, mass M and moment of inertia

I

rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is

a = {{g\,\sin \,\theta } \over {1 + {I \over {M{R^2}}}}}