Circular Motion Questions — JEE Physics

Q1

A particle is moving with constant speed in a circular path. When the particle turns by an angle

90^{\circ}

, the ratio of instantaneous velocity to its average velocity is

\pi: x \sqrt{2}

. The value of

x

will be -

A 1

B 7

C 5

D 2

Correct Answer

Option D

Solution

\begin{aligned} & \text{ Instantaneous velocity }=\omega R \\\\ & \text{ Time taken }=\frac{\pi}{2 \omega} \\\\ & \text{ Displacement }=R \sqrt{2} \\\\ & \text{ Average velocity }=\frac{R \sqrt{2} \times 2 \omega}{\pi}=\frac{2 \sqrt{2}}{\pi} \omega R \\\\ & \Rightarrow \frac{v_{\text{ins }}}{v_{\text{avg }}}=\frac{\omega R \pi}{2 \sqrt{2} \omega R} \\\\ & \Rightarrow x=2 \end{aligned}

Q2

A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length

\ell

. The other end is fixed. The system is given an angular speed

\omega

about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is :

A

{{m\ell {\omega ^2}} \over {k - m{\omega ^2}}}

B

{{m\ell {\omega ^2}} \over {k - m{\omega}}}

C

{{m\ell {\omega ^2}} \over {k + m{\omega ^2}}}

D

{{m\ell {\omega ^2}} \over {k + m{\omega}}}

Correct Answer

Option A

Solution

At elongated position (x), Fradial = mr

{\omega ^2}

$\therefore$ kx = m

\left( {{l} + x} \right)

{\omega ^2}

$\Rightarrow$ x =

{{m\ell {\omega ^2}} \over {k - m{\omega ^2}}}

Q3

A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by the side walls of the groove is :

A 9.859

\times

10

-

2 N

B 0.0314 N

C 9.859

\times

10

-

4 N

D 6.28

\times

10

-

3 N

Correct Answer

Option C

Solution

Normal force will provide the necessary centripetal force. $\Rightarrow$ N = m $\omega$ 2R Also, $\omega$ =

{{2\pi } \over t}

N = (0.2)

\left( {{{4{\pi ^2}} \over {{T^2}}}} \right)

(0.2) $\Rightarrow$ N = 0.2 $\times$

{{4 \times {{(3.14)}^2}} \over {{{(40)}^2}}}

$\times$ 0.2 $\therefore$ N = 9.859 $\times$ 10 $-$ 4 N

Q4

A car is moving on a horizontal curved road with radius 50 m. The approximate maximum speed of car will be, if friction between tyres and road is 0.34. [take g = 10 ms

^{-2}

]

A 3.4 ms

^{-1}

B 13 ms

^{-1}

C 22.4 ms

^{-1}

D 17 ms

^{-1}

Correct Answer

Option B

Solution

$f_{s}=\dfrac{m v^{2}}{r}$ For maximum speed in safe turning, $\mathrm{f}_{\mathrm{s}}=\mathrm{f}_{\mathrm{s}} \max =\mu \mathrm{mg}$ $\mathrm{v}_{\max }$ (for safe turning) $=\sqrt{\mu \mathrm{rg}}$ $=\sqrt{0.34 \times 50 \times 10} \approx 13 \mathrm{~m} / \mathrm{s}$

Q5

The minimum velocity (in

m{s^{ - 1}}

) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction

0.6

to avoid skidding is

A

60

B

30

C

15

D

25

Correct Answer

Option B

Solution

For no skidding along curved track, The maximum velocity possible

{v_{\max }} = \sqrt {\mu rg}

Here

\mu = 0.6,\,r = 150m,\,g = 9.8

$\therefore$

{v_{\max }} = \sqrt {0.6 \times 150 \times 9.8} \simeq 30m/s

Q6

Which of the following statements is FALSE for a particle moving in a circle with a constant angular speed?

A The velocity vector is tangent to the circle.

B The acceleration vector is tangent to the circle.

C The acceleration vector points to the centre of the circle.

D The velocity and acceleration vectors are perpendicular to each other.

Correct Answer

Option B

Solution

Only option

(b)

is false since acceleration vector acts along the radius of the circle or towards center of the circle for uniform circular motion and velocity vector always acts along the tangent of the circle.

Q7

Two cars of masses m1 and m2 are moving in circles of radii r1 and r2, respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is

A m1r1 : m2r2

B m1 : m2

C r1 : r2

D 1 : 1

Correct Answer

Option C

Solution

We know,

a = r\,{w^2} = r \times {\left( {{{2\pi } \over T}} \right)^2}

Given,

{T_1} = {T_2} = T

{a_1} = {r_1} \times {\left( {{{2\pi } \over T}} \right)^2}

{a_2} = {r_2} \times {\left( {{{2\pi } \over T}} \right)^2}

$\therefore$

{{{a_1}} \over {{a_2}}} = {{{r_1}} \over {{r_2}}}

Q8

A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of

3.5

revolutions per second. A coin placed at a distnce of 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is : (g = 10 m/s2)

A 0.5

B 0.3

C 0.7

D 0.6

Correct Answer

Option D

Solution

We have

m{\omega ^2}r = \mu mg

......

(1) Given : rate of rotation = 3.5 rev/s $\Rightarrow$ 1 revolution = 2 $\pi$ rad That is, 3.5 revolutions = 3.5 $\times$ 2 $\pi$ rad Therefore, $\omega$ = 3.5 $\times$ 2 $\pi$ rad/s r = 1.25 cm = 1.25 $\times$ 10 $-$ 2 m Thus, from Eq. (1), we have

m{\omega ^2}r = \mu mg \Rightarrow {\omega ^2}r = \mu g

\Rightarrow \mu = {{{\omega ^2}r} \over g} = {{{{(3.5 \times 2\pi )}^2}(1.25 \times {{10}^{ - 2}})} \over {10}}

\Rightarrow \mu = 0.60

Q9

Statement I : A cyclist is moving on an unbanked road with a speed of 7 kmh

-

1 and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s2) Statement II : If the road is banked at an angle of 45

^\circ

, cyclist can cross the curve of 2m radius with the speed of 18.5 kmh

-

1 without slipping. In the light of the above statements, choose the correct answer from the options given below.

A Statement I is incorrect and statement II is correct

B Both statement I and statement II are true

C Statement I is correct and statement II is incorrect

D Both statement I and statement II are false

Correct Answer

Option B

Solution

On a horizontal ground,

{v_{\max }} = \sqrt {\mu Rg} = \sqrt {0.2 \times 2 \times 9.8} = 1.97

m/s =

1.97 \times {{18} \over 5}

= 7.12 km/hr = 7.2 km/hr Statement - 2

{v_{\max }} = \sqrt {gr\left( {{{\tan \theta + \mu } \over {1 - \mu \tan \theta }}} \right)} = \sqrt {2 \times 9.8 \times {{12} \over {0.8}}} = 19.5

km/hr

{v_{\min }} = \sqrt {rg\left( {{{\tan \theta - \mu } \over {1 + \mu \tan \theta }}} \right)} = \sqrt {2 \times 9.8 \times {{0.8} \over {1.2}}} = 12.01

km/hr

Q10

The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a 30

^\circ

banked road at maximum possible speed without skidding is ____________

\times

103 kg m/s2. [Given cos30

^\circ

= 0.87,

\mu

s = 0.2]

A 12.4

B 7.2

C 6.96

D 10.2

Correct Answer

Option D

Solution

The given situation can be represented as Equating forces perpendicular to the inclined plane,

N = mg\cos 30^\circ + {{m{v^2}} \over R}\sin 30^\circ

\Rightarrow N - mg\cos 30^\circ = {{m{v^2}} \over R}\sin 30^\circ

.... (i) Equating forces along the inclined plane,

mg\sin 30^\circ + {\mu _s}N = {{m{v^2}} \over R}\cos 30^\circ

.... (ii) On dividing Eq. (i) by Eq. (ii), we get

{{N - mg\cos 30^\circ } \over {mg\sin 30^\circ + {\mu _s}N}} = \tan 30^\circ

[ $\because$

\cos 30^\circ = {{\sqrt 3 } \over 2}

and

\sin 30^\circ = {1 \over 2}

] $\Rightarrow$

{{N - mg(\sqrt3/2) } \over {mg(1/2) + (0.2)N}} = {1 \over {\sqrt 3 }}

N\sqrt 3 - mg{{\sqrt 3 } \over 2}.\sqrt 3 = {{mg} \over 2} + 0.2N

\Rightarrow (\sqrt 3 - 0.2)N = mg{{(1 + 3)} \over 2} = 2mg

\Rightarrow N = {{2mg} \over {\sqrt 3 - 0.2}} = {{2 \times 800 \times 10} \over {1.532}}

= 10.44 \times {10^3}V

Therefore,

N = 10.2 \times {10^3}

kg-m/s2