Rotation The Laws(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

ROTATION THE LAWS MCQs

Total Questions : 28 | Page 3 of 3 pages

Question 21. A uniform solid cylinder of mass 'M' and radius 'R' is resting on a horizontal platform (which is parallel to X-Y plane) with its axis along the Y-axis and free to roll on the platform. The platform is given a motion in X-direction given by

x = A cos ω t

. There is no slipping between the cylinder and the platform. The maximum torque acting on the cylinder as measured about its centre of mass is

12MRAω2
MRAω2
2mRAω2
mRAωA2cos2ωt

Discuss Question

Answer: Option A. -> 12MRAω2
:
A

⃗ τ = I α = \frac{1}{2} M R^{2} \times (\frac{L i n e a r a c c e l e r a t i o n}{R})

acceleration =

ω^{2} A sin ω t

(l i n e a r a c c e l e r a t i o n)_{m a x} = ω^{2} A

∴ (τ)_{m a x} = \frac{1}{2} \frac{M R^{2} ω^{2}}{R} = \frac{1}{2} M R A ω^{2} .

Question 22. The angular velocity of a body is

⃗ ω = 2 ˆ i + 3 ˆ j + 4 ˆ k

and a torque

⃗ τ = ˆ i + 2 ˆ j + 3 ˆ k

acts on it. The rotational power will be

20 W
15 W
√17W
√14W

Discuss Question

Answer: Option A. -> 20 W
:
A
Power

(P) = ⃗ τ . ⃗ ω = (i + 2 ˆ j + 3 ˆ k) . (2 ˆ i + 3 ˆ j + 4 ˆ k) = 2 + 6 + 12 = 20 W

Question 23. A centrifuge consists of four solid cylindrical containers, each of mass 'm' at radial distance 'r' from the axis of rotation. A time 't' is required to bring the centrifuge to an angular velocity

ω

from rest under a constant torque

τ

applied to the shaft. The radius of each container is 'a' and the mass of the shaft and the supporting arms is small compared to 'm'. Then

A Centrifuge Consists Of Four Solid Cylindrical Containers, ...

t=4mr2ωτ
t=2m(a2+2r2)ωτ
t=2ma2ωτ
t=2m(r2+2a2)ωτ

Discuss Question

Answer: Option B. -> t=2m(a2+2r2)ωτ
:
B

L = τ t

\Rightarrow t = \frac{4 (m r^{2} + \frac{1}{2} m a^{2}) ω}{τ}

\Rightarrow t = \frac{2 m (a^{2} + 2 r^{2}) ω}{τ}

Question 24. A disc of mass 'M' and radius 'R' is rolling with an angular speed of

ω

rad/s on a horizontal plane as shown in the figure. The magnitude of angular momentum of the disc about the origin O is

A Disc Of Mass 'M' And Radius 'R' Is Rolling With An Angular...

(12)MR2ω
MR2ω
(32)MR2ω
2MR2ω

Discuss Question

Answer: Option C. -> (32)MR2ω
:
C
The angular momentum of a body

⃗ L

may be expressed as the sum of two parts,
(a) one arising from the motion of the centre of mass of the body

({⃗ L}_{o r b i t a l})

(b) the other from the motion of the body with respect to its centre of mass

({⃗ L}_{s p i n})

i.e.,

{⃗ L}_{t o t a l} = {⃗ L}_{C . M} + {⃗ r}_{C . M} \times ⃗ p

\Rightarrow {⃗ L}_{t o t a l} = {⃗ L}_{C . M .} + M (\to C . M \times {⃗ v}_{C . M})

For this problem

L_{C . M} = I ω = \frac{1}{2} M R^{2} ω

and

M ({⃗ r}_{C . M} \times {⃗ v}_{C . M}) = M R v_{C M} = M R (R ω)

\Rightarrow M ({⃗ r}_{C . M} \times {⃗ v}_{C . M .}) = M R^{2} ω

\Rightarrow L_{t o t a l} = \frac{1}{2} M R^{2} ω + M R^{2} ω = \frac{3}{2} M R^{2} ω

Question 25. A uniform solid cylinder of mass 'm' can rotate freely about its axis which is kept horizontal. A particle of mass

m_{0}

hangs from the end of a light string wound round the cylinder. When the system is allowed to move, the acceleration with which the particle descends is

2m0gm0+2m
2m0gm+2m0
m0gm+m0
2m0gm+2m0

Discuss Question

Answer: Option B. -> 2m0gm+2m0
:
B
Suppose that the tension in the string is T.
Then,

= m_{0} g - T = m_{0} a

∴ T = m_{0} (g - a)

where a = acceleration
Further, T.r = I

α

α

=angular acceleration and T.r = moment of force acting on cylinder
or

T = \frac{I α}{r}

= (\frac{m r^{2}}{2}) (\frac{α}{r}) = \frac{m r α}{2} = \frac{m a}{2}

∴ a = \frac{2 T}{m} = \frac{2 m_{0} (g - a)}{m}

Solving for a, we get

a = (\frac{2 m_{0} g}{m + 2 m_{0}}) .

Question 26. A uniform circular disc of radius 'R' with a concentric circular hole of radius

\frac{R}{2}

rolls on a horizontal plane. The fraction of its total energy associated with its rotational motion is

Discuss Question

Answer: Option B. -> 513
:
B

I = \frac{5}{8} m r^{2}

\frac{R . K . E}{T . E} = \frac{1}{1 + \frac{R^{2}}{K^{2}}} = \frac{1}{1 + \frac{8}{5}} = \frac{5}{13}

Question 27. A solid body rotates with deceleration about a stationary axis with an angular deceleration

| α | = k \sqrt{ω}

; where 'k' is a constant and

ω

is the angular velocity of the body. If the initial angular velocity is

ω_{0}

, then mean angular velocity of the body averaged over the whole time of rotation is

ω0
ω02
ω03
ω04

Discuss Question

Answer: Option C. -> ω03
:
C

- \frac{d ω}{d t} = k ω^{1 / 2}

\Rightarrow - \int_{ω_{0}}^{ω} \frac{d ω}{ω^{1 / 2}} = k \int_{0}^{t} d t

\Rightarrow - 2 {[\sqrt{ω}]}_{ω_{0}}^{ω} = k t

\Rightarrow 2 \sqrt{ω_{0}} - 2 \sqrt{ω} = k t

\Rightarrow 2 \sqrt{ω} = 2 \sqrt{ω_{0}} - k t; ω = {(\sqrt{ω_{0}} - \frac{1}{2} k t)}^{2}

The body comes to a stop

(ω = 0)

in time

t_{0} = \frac{2 \sqrt{ω_{0}}}{k}

ω = {(\sqrt{ω_{0}} - \frac{k t}{2})}^{2}

\Rightarrow ω_{=} ω_{0} + \frac{k^{2} t^{2}}{4} - k t \sqrt{ω_{0}}

Since

< ω_{a r} >= \frac{\int_{0}^{t_{0}} ω d t}{\int_{0}^{0} d t} = \frac{ω_{0} t_{0} + \frac{k^{2}}{4} \frac{t_{0}^{3}}{3} - k \sqrt{ω_{0}} \frac{t_{0}^{2}}{2}}{t_{0}}

\Rightarrow< ω_{a v} >= ω_{0} + \frac{k^{2}}{4} \frac{t_{0}^{2}}{3} - k \sqrt{ω_{0}} . \frac{t_{0}}{2}

\Rightarrow< ω_{a r} >= ω_{0} + \frac{k^{2}}{12} \frac{4 ω_{0}}{k^{2}} - \frac{k \sqrt{ω_{0}} 2 \sqrt{ω_{0}}}{2 k}

\Rightarrow< ω_{a v} >= ω_{0} + \frac{ω_{0}}{3} - ω_{0}

\Rightarrow< ω_{a v} >= \frac{ω_{0}}{3} .

Question 28. A horizontal heavy uniform bar of weight 'W' is supported at its ends by two men. At the instant, one of the men lets go off his end of the rod, the other feels the force on his hand changed to

Discuss Question

Answer: Option D. -> W4
:
D
Let the mass of the rod is M