Rotational Motion — Page 4 | NEET Physics

Q31

A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s

-

2 . Its net acceleration in m s

-

2 at the end of 2.0 s is approximately

A 6.0

B 3.0

C 8.0

D 7.0

Correct Answer

Option C

Solution

Given, r = 50 cm = 0.5 m, $\alpha$ = 2.0 rad s –2 ,

{\omega _0}

= 0 At the end of 2 s, Tangential acceleration, at = r $\alpha$ = 0.5 × 2 = 1 m s –2 Radial acceleration, =

{a_r} = {\omega ^2}r = {\left( {{\omega _0} + \alpha t} \right)^2}r

= (0 + 2 × 2) 2 × 0.5 = 8 m s –2 $\therefore$ Net acceleration,

a = \sqrt {a_t^2 + a_r^2} = \sqrt {{1^2} + {8^2}} = \sqrt {65} \approx 8\,m{s^{ - 2}}

Q32

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

A 11 MR 2 /32

B 9 MR 2 /32

C 15 MR 2 /32

D 13 MR 2 /32

Correct Answer

Option D

Solution

Moment of inertia of complete disc about point 'O'.

{I_{Total\,disc}} = {{M{R^2}} \over 2}

Mass of removed disc

{M_{{\mathop{\rm Removed}\nolimits} }} = {M \over 4}\left( {Mass \propto area} \right)

Moment of inertia of removed disc about point 'O'. I Removed (about same perpendicular axis)

= {I_{cm}} + m{x^2}

= {M \over 4}{{{{\left( {R/2} \right)}^2}} \over 2} + {M \over 4}{\left( {{R \over 2}} \right)^2} = {{3M{R^2}} \over {32}}

Therefore the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the centre, I Remaing disc = I Total – I Removed

= {{M{R^2}} \over 2} - {3 \over {32}}M{R^2} = {{13} \over {32}}M{R^2}

Q33

A force

\overrightarrow F = \alpha \widehat i + 3\widehat j + 6\widehat k

is acting at a point

\overrightarrow r = 2\widehat i - 6\widehat j - 12\widehat k

. The value of

\alpha

for which angular momentum about origin is conserved is

A zero

B 1

C

-

1

D 2

Correct Answer

Option C

Solution

From Newton's second law for rotational motion,

\overrightarrow \tau = {{\overrightarrow d L} \over {dt}}

, if

\overrightarrow L

= constant then

\overrightarrow \tau = 0

So,

\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = 0

\left( {2\widehat i - 6\widehat j - 12\widehat k} \right) \times \left( {\alpha \widehat i + 3\widehat j + 6\widehat k} \right) = 0

Solving we get $\alpha$ = –1

Q34

An automobile moves on a road with a speed of 54 km h

-

1 . The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m 2 . If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its brakes to the wheel is

A 10.86 kg m 2 s

-

2

B 2.86 kg m 2 s

-

2

C 6.66 kg m 2 s

-

2

D 8.58 kg m 2 s

-

2

Correct Answer

Option C

Solution

Here, Speed of the automobile,

v = 54\,km{h^{ - 1}} = 54 \times {5 \over {18}}m{s^{ - 1}} = 15m{s^{ - 1}}

Radius of the wheel of the automobile, R = 0.45 m Moment of inertia of the wheel about its axis of rotation, I = 3 kg m 2 Time in which the vehicle brought to rest, t = 15 s The initial angular speed of the wheel is

{\omega _i} = {v \over R} = {{15m{s^{ - 1}}} \over {0.45\,m}} = {{1500} \over {45}}rad\,{s^{ - 1}} = {{100} \over 3}rad\,{s^{ - 1}}

and its final angular speed is

{\omega _f}

= 0 (as the vehicle comes to rest) $\therefore$ The angular retardation of the wheel is

\alpha = {{{\omega _f} - {\omega _i}} \over t} = {{0 - {{100} \over 3}} \over {15\,s}} = - {{100} \over {45}}rad\,{s^{ - 1}}

The magnitude of required torque is

\tau = I\left| \alpha \right| = \left( {3kg\,{m^2}} \right)\left( {{{100} \over {45}}rad\,{s^{ - 2}}} \right)

= {{20} \over 3}kg\,{m^2}{s^{ - 2}} = 6.66\,kg\,{m^2}{s^{ - 2}}

Q35

Point masses m 1 and m 2 are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity

\omega

0 is minimum, is given by

A

x = {{{m_2}} \over {{m_1}}}L

B

x = {{{m_2}L} \over {{m_1} + {m_2}}}

C

x = {{{m_1}L} \over {{m_1} + {m_2}}}

D

x = {{{m_1}} \over {{m_2}}}L

Correct Answer

Option B

Solution

K = {1 \over 2}\left( {{m_1}{x^2}} \right)\omega _0^2 + {1 \over 2}{m_2}{\left( {L - x} \right)^2}\omega _0^2

dK/dx = 0

{m_1}x = {m_2}\left( {L - x} \right)

{m_1}x = {m_2}L - {m_2}x

\Rightarrow x = {{{m_2}L} \over {{m_1} + {m_2}}}

Q36

A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

A

{{W\left( {d - x} \right)} \over x}

B

{{W\left( {d - x} \right)} \over d}

C

{{Wx} \over d}

D

{{Wd} \over x}

Correct Answer

Option B

Solution

Given situation is shown in figure.

N 1 = Normal reaction on A N 2 = Normal reaction on B W = Weight of the rod In vertical equilibrium, N 1 + N 2 = W …(i) Torque balance about centre of mass of the rod, N 1 x = N 2 (d – x) Putting value of N 2 from equation (i) N 1 x = (W – N 1 )(d – x) $\Rightarrow$ N 1 x = Wd – Wx – N 1 d + N 1 x $\Rightarrow$ N 1 d = W(d – x) $\therefore$

{N_1} = {{W\left( {d - x} \right)} \over d}

Q37

A mass m moves in a circle on a smooth horizontal plane with velocity v 0 at a radius R 0 . The mass is attached to a string which passes through a smooth hole in the plane as shown. the tension in the string is increased gradually and finally m moves in a circle of radius

{{{R_0}} \over 2}

. The final value of the kinetic energy is

A 2mv

_0^2

B

{1 \over 2}

mv

_0^2

C mv

_0^2

D

{1 \over 4}

mv

_0^2

Correct Answer

Option A

Solution

According to law of conservation of angular momentum mvr = mv'r'

{v_0}{R_0} = v\left( {{{{R_0}} \over 2}} \right);v = 2{v_0}

…(i) $\therefore$

{{{K_0}} \over K} = {{{1 \over 2}mv_0^2} \over {{1 \over 2}m{v^2}}} = {\left( {{{{v_0}} \over v}} \right)^2}

$\Rightarrow$

{K \over {{K_0}}} = {\left( {{v \over {{v_0}}}} \right)^2} = {\left( 2 \right)^2}

(Using (i)) K = 4K 0 = 2

{mv_0^2}

Q38

Three identical spherical shells, each of mass m and radius r are placed as shown in figure. Consider an axis XX' which is touching to two shells and passing through diameter of third shell. Moment of inertia of the system consisting of these three spherical shells about XX' axis is

A

{{16} \over 5}m{r^2}

B 4mr 2

C

{{11} \over 5}m{r^2}

D 3mr 2

Correct Answer

Option B

Solution

Net moment of inertia of the system, I = I 1 + I 2 + I 3 The moment of inertia of a shell about its diameter, I 1 =

{2 \over 3}

mr 2 The moment of inertia of a shell about its tangent is given by

{I_2} = {I_3} = {I_1} + m{r^2} = {2 \over 3}m{r^2} + m{r^2} = {5 \over 3}m{r^2}

$\therefore$

I = 2 \times {5 \over 3}m{r^2} + {2 \over 3}m{r^2} = {{12m{r^2}} \over 3} = 4m{r^2}

Q39

A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions s

-

2 is

A 25 N

B 50 N

C 78.5 N

D 157 N

Correct Answer

Option D

Solution

Here $\alpha$ = 2 revolutions/s 2 = 4 $\pi$ rad/s 2 (given)

{I_{cylinder}} = {1 \over 2}M{R^2} = {1 \over 2}\left( {50} \right){\left( {0.5} \right)^2} = {{25} \over 4}{\rm{Kg - }}{{\rm{m}}^{\rm{2}}}

As

\tau = I\alpha

so

TR = I\alpha

\Rightarrow T = {{I\alpha } \over R} = {{\left( {{{25} \over 4}} \right)\left( {4\pi } \right)} \over {0.5}}N

= 50 $\pi$ N = 157 N

Q40

The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle

\theta

without slipping and slipping down the incline without rolling is

A 5 : 7

B 2 : 3

C 2 : 5

D 7 : 5

Correct Answer

Option A

Solution

Acceleration of the solid sphere slipping down the incline without rolling is

{a_{slipping}} = g\sin \theta

....(i) Acceleration of the solid sphere rolling down the incline without slipping is

{a_{rolling}} = {{g\sin \theta } \over {1 + {{{k^2}} \over {{R^2}}}}} = {{g\sin \theta } \over {1 + {2 \over 5}}}

\left( \because{For\,\,solid\,\,sphere,{{{k^2}} \over {{R^2}}} = {2 \over 5}} \right)

=

{5 \over 7}g\sin \theta

... (ii) Divide eqn. (ii) by eqn. (i), we get

{{{a_{rolling}}} \over {{a_{slipping}}}} = {5 \over 7}