Circular Motion — Page 4 | JEE Physics

Q31

A particle is moving along a circular path with a constant speed of 10 ms–1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60o around the centre of the circle?

A zero

B 10 m/s

C

10\sqrt 2 m/s

D

10\sqrt 3 m/s

Correct Answer

Option B

Solution

\left| {\Delta \overrightarrow v } \right| = \sqrt {v_1^2 + v_2^2 + 2{v_1}{v_2}\cos \left( {\pi - \theta } \right)}

= 2v\sin {\theta \over 2}

since

\left[ {\left| {\overline v {}_1} \right| = \left| {{{\overline v }_2}} \right|} \right]

= \left( {2 \times 10} \right) \times \sin \left( {{{30}^o}} \right)

= 10 m/s

Q32

A car of mass '

m

' moves on a banked road having radius '

r

' and banking angle

\theta

. To avoid slipping from banked road, the maximum permissible speed of the car is

v_0

. The coefficient of friction

\mu

between the wheels of the car and the banked road is

A

\mu=\dfrac{v_0^2+r g \tan \theta}{r g+v_0^2 \tan \theta}

B

\mu=\dfrac{v_0^2-r g \tan \theta}{\mathrm{rg}-\mathrm{v}_{\mathrm{o}}^2 \tan \theta}

C

\mu=\dfrac{v_0^2-r g \tan \theta}{r g+v_0^2 \tan \theta}

D

\mu=\dfrac{v_o^2+r g \tan \theta}{r g-v_o^2 \tan \theta}

Correct Answer

Option C

Solution

N \cos\theta - \mu N \sin\theta = mg

N \sin\theta + \mu N \cos\theta = \frac{mv_0^2}{r}

Dividing the second equation by the first gives

\frac{v_0^2}{r} = \frac{g\left(\sin\theta + \mu \cos\theta\right)}{\cos\theta - \mu \sin\theta}.

Multiplying both sides by $\cos\theta - \mu\sin\theta$ leads to

\frac{v_0^2}{r}(\cos\theta - \mu \sin\theta) = g(\sin\theta + \mu \cos\theta).

Expanding and grouping the terms with $\mu$ :

\frac{v_0^2}{r}\cos\theta - g\sin\theta = \mu\left(\frac{v_0^2}{r}\sin\theta + g\cos\theta\right).

Thus, solving for $\mu$ :

\mu = \frac{\frac{v_0^2}{r}\cos\theta - g\sin\theta}{\frac{v_0^2}{r}\sin\theta + g\cos\theta}.

Multiplying the numerator and the denominator by $r$ gives

\mu = \frac{v_0^2\cos\theta - rg\sin\theta}{v_0^2\sin\theta + rg\cos\theta}.

Dividing both the numerator and denominator by $\cos\theta$ (assuming $\cos\theta \neq 0$ ):

\mu = \frac{v_0^2 - rg\tan\theta}{v_0^2\tan\theta + rg}.

This result matches the expression

\mu=\frac{v_0^2 - rg\,\tan\theta}{rg + v_0^2\,\tan\theta}.

Thus, the correct answer is: Option C

Q33

If the radius of curvature of the path of two particles of same mass are in the ratio

3: 4

, then in order to have constant centripetal force, their velocities will be in the ratio of :

A

1: \sqrt{3}

B

2: \sqrt{3}

C

\sqrt{3}: 2

D

\sqrt{3}: 1

Correct Answer

Option C

Solution

Given

\mathrm{m}_1=\mathrm{m}_2

\text{ and } \frac{r_1}{r_2}=\frac{3}{4}

As centripetal force

\mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}}

In order to have constant (same in this question) centripetal force

\begin{aligned} & \mathrm{F}_1=\mathrm{F}_2 \\ & \frac{\mathrm{m}_1 \mathrm{v}_1^2}{\mathrm{r}_1}=\frac{\mathrm{m}_2 \mathrm{v}_2^2}{\mathrm{r}_2} \\ & \Rightarrow \frac{\mathrm{v}_1}{\mathrm{v}_2}=\sqrt{\frac{\mathrm{r}_1}{\mathrm{r}_2}}=\frac{\sqrt{3}}{2} \end{aligned}

Q34

A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r such that

r = {L \over {\sqrt 2 }}

. The speed of particle will be :

A

{\sqrt {rg} }

B

{\sqrt {2rg} }

C

{2\sqrt {rg} }

D

{\sqrt {{{rg} \over 2}} }

Correct Answer

Option A

Solution

r = {l \over {\sqrt 2 }}

\sin \theta = {r \over l} = {l \over {\sqrt 2 }}

$\theta$ = 45

^\circ

T\sin \theta = {{m{v^2}} \over r}

T\cos \theta = mg

\tan \theta = {{{v^2}} \over {rg}} \Rightarrow v = \sqrt {rg}

Q35

For a particle in uniform circular motion the acceleration

\overrightarrow a

at a point P(R, θ) on the circle of radius R is (here θ is measured from the x–axis)

A

- {{{v^2}} \over R}\cos \theta \widehat i + {{{v^2}} \over R}\sin \theta \widehat j

B

- {{{v^2}} \over R}\sin \theta \widehat i + {{{v^2}} \over R}\cos \theta \widehat j

C

- {{{v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j

D

{{{v^2}} \over R}\widehat i + {{{v^2}} \over R}\widehat j

Correct Answer

Option C

Solution

For a particle in uniform circular motion,

{a_c} = {{{v^2}} \over R}

towards the center of the circle From figure,

\overrightarrow a = {a_c}\cos \theta \left( { - \widehat i} \right) + {a_c}\sin \theta \left( { - \widehat j} \right)

= {{ - {v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j

Q36

A particle is moving with uniform speed along the circumference of a circle of radius R under the action of a central fictitious force F which is inversely proportional to R3. Its time period of revolution will be given by :

A

T \propto {R^{{4 \over 3}}}

B

T \propto {R^{{5 \over 2}}}

C

T \propto {R^{{3 \over 2}}}

D

T \propto {R^2}

Correct Answer

Option D

Solution

F \propto {1 \over {{R^3}}}

F = {K \over {{R^3}}}

{{m{v^2}} \over R} = {K \over {{R^3}}}

m{(\omega R)^2} = {K \over {{R^2}}}

m{\omega ^2}{R^2} = {K \over {{R^2}}}

{\omega ^2} = {K \over m}\left( {{1 \over {{R^4}}}} \right)

{\left( {{{2\pi } \over T}} \right)^2} \propto {1 \over {{R^4}}}

{{4{\pi ^2}} \over {{T^2}}} \propto {1 \over {{R^4}}}

T \propto {R^2}

Q37

A coin is placed on a disc. The coefficient of friction between the coin and the disc is

\mu

. If the distance of the coin from the center of the disc is

r

, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is :

A

\sqrt{\dfrac{r}{\mu g}}

B

\sqrt{\dfrac{\mu g}{r}}

C

\dfrac{\mu g}{r}

D

\dfrac{\mu}{\sqrt{r g}}

Correct Answer

Option B

Solution

When the coin is on the disc and the disc starts rotating, centrifugal force acts on the coin, trying to push it away from the center.

The friction between the coin and the disc opposes this motion.

For the coin to not slip, the frictional force must be equal to the centrifugal force acting on the coin.The normal force (N) acting on the coin is equal to the weight of the coin, which can be represented as:

N = mg

where

m

is the mass of the coin, and

g

is the acceleration due to gravity.The centrifugal force (

f

) that acts on the coin due to the rotation of the disc is given by:

f = m\omega^2r

where $\omega$ is the angular velocity, and

r

is the distance of the coin from the center of the disc.Friction force (f) is also given by the formula:

f = \mu N

where $\mu$ is the coefficient of static friction between the coin and the disc.For the coin to not slip, the centrifugal force must be equal to the frictional force, hence:

\mu mg = m\omega^2r

Dividing both sides by

mr

, we get:

\omega = \sqrt{\frac{\mu g}{r}}

This equation shows that the maximum angular velocity ( $\omega$ ) that can be given to the disc to prevent the coin from slipping off depends on the coefficient of friction ( $\mu$ ), the acceleration due to gravity (

g

), and the distance of the coin from the center (

r

).

Q38

A car of

800 \mathrm{~kg}

is taking turn on a banked road of radius

300 \mathrm{~m}

and angle of banking

30^{\circ}

. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely:

(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sqrt{3}=1.73)

A 51.4 m/s

B 102.8 m/s

C 70.4 m/s

D 264 m/s

Correct Answer

Option A

Solution

When a car takes a turn on a banked road, the forces involved are the gravitational force, the normal force from the surface, and frictional force (if any).

The net force provides the necessary centripetal force for the circular motion.

The angle of banking and static friction contribute to the maximum speed the car can achieve without slipping.

The forces acting on the car are as follows: 1.

The normal force (

N

) acts perpendicular to the surface of the road. 2. Gravitational force (

mg

) acts downward. 3. Frictional force (

f

), which can provide additional centripetal force if needed.

It acts parallel to the surface of the road, towards the center of the circle.

The normal force and the gravitational force components can be resolved into two directions: perpendicular and parallel to the road surface.

The maximum speed is achieved when all available forces (normal, frictional) are utilized to provide the necessary centripetal force (

F_c

) without slipping. The centripetal force required for circular motion is given by: $F_c = \dfrac{mv^2}{r}$ Where:

m

= mass of the car =

800 kg

v

= speed of the car

r

= radius of the turn =

300 m

On a banked curve, the maximum velocity can be calculated using the formula: $v = \sqrt{rg(\tan\theta + \mu)\Big/\big(1-\mu\tan\theta)}$ Where: $\theta$ = angle of banking =

30^{\circ}

$\mu$ = coefficient of static friction =

0.2

g

= acceleration due to gravity =

10 m/s^2

First, calculate the tangent of the angle: $\tan30^{\circ} = \dfrac{1}{\sqrt{3}} = \dfrac{1}{1.73}$ Substitute all the values into the formula: $v = \sqrt{300 \times 10 \times \left(\dfrac{1}{1.73} + 0.2\right)\Big/\big(1 - 0.2 \times \dfrac{1}{1.73}\big)}$ $v = \sqrt{3000 \times \left(\dfrac{1}{1.73} + 0.2\right)\Big/\big(1 - \dfrac{0.2}{1.73}\big)}$ $v = \sqrt{3000 \times \dfrac{1 + 1.73 \times 0.2}{1.73 - 0.2}}$ $v = \sqrt{3000 \times \dfrac{1 + 0.346}{1.73 - 0.2}}$ $v = \sqrt{3000 \times \dfrac{1.346}{1.53}}$ $v = \sqrt{3000 \times 0.8797}$ $v = \sqrt{2639.1} \approx 51.37 \, m/s$ Hence, the closest option to the calculated maximum speed without slipping is: Option A: 51.4 m/s

Q39

A stone of mass

1 \mathrm{~kg}

is tied to end of a massless string of length

1 \mathrm{~m}

. If the breaking tension of the string is

400 \mathrm{~N}

, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is :

A

20 \mathrm{~ms}^{-1}

B

40 \mathrm{~ms}^{-1}

C

400 \mathrm{~ms}^{-1}

D

10 \mathrm{~ms}^{-1}

Correct Answer

Option A

Solution

\begin{aligned} & T \sin \theta=\frac{m v^{2}}{l \sin \theta} \\\\ & \cos \theta=\frac{m g}{T} ........(1) \\\\ & \sin^2 \theta=\frac{m v^{2}}{T l} ........(2) \end{aligned}

From (1) and (2), $1=\left(\dfrac{m g}{T}\right)^{2}+\dfrac{m v^{2}}{T l}$ $\Rightarrow$ $1=\left(\dfrac{10}{400}\right)^{2}+\dfrac{v^{2}}{400}$ $\Rightarrow$ $v^{2}=399.78$ $\Rightarrow$ $v=20 \mathrm{~m} / \mathrm{s}$

Q40

A body of mass ‘m’ connected to a massless and unstretchable string goes in vertical circle of radius ‘R’ under gravity g. The other end of the string is fixed at the center of circle. If velocity at top of circular path is

n\sqrt{ g R}

, where, n ≥ 1, then ratio of kinetic energy of the body at bottom to that at top of the circle is :

A

\dfrac{n^2 + 4}{n^2}

B

\dfrac{n + 4}{n}

C

\dfrac{n^2}{n^2 + 4}

D

\dfrac{n}{n + 4}

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{V}_{\text{Top }}=\sqrt{\mathrm{n}^2 \mathrm{gR}} \\ & \mathrm{~V}_{\text{Botom }}=\sqrt{\mathrm{n}^2 \mathrm{gR}+4 \mathrm{gR}} \\ & \text{ Ratio }=\frac{\mathrm{n}^2+4}{\mathrm{n}^2} \end{aligned}