Circular Motion — Page 2 | JEE Physics

Q11

A huge circular arc of length 4.4 ly subtends an angle '4s' at the centre of the circle. How long it would take for a body to complete 4 revolution if its speed is 8 AU per second? Given : 1 ly = 9.46

\times

1015 m 1 AU = 1.5

\times

1011 m

A 4.1

\times

108 s

B 4.5

\times

1010 s

C 3.5

\times

106 s

D 7.2

\times

108 s

Correct Answer

Option B

Solution

R = {l \over \theta }

Time

= {{4 \times 2\pi R} \over v} = {{4 \times 2\pi } \over v}\left( {{l \over \theta }} \right)

put l = 4.4 $\times$ 9.46 $\times$ 1015 v = 8 $\times$ 1.5 $\times$ 1011

\theta = {4 \over {3600}} \times {\pi \over {180}}

rad. we get time = 4.5 $\times$ 1010 sec.

Q12

A stone tide to a spring of length L is whirled in a vertical circle with the other end of the spring at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is

\sqrt {x({u^2} - gL)}

. The value of x is -

A 3

B 2

C 1

D 5

Correct Answer

Option B

Solution

\overrightarrow v = \sqrt {{u^2} - 2gL} \widehat j

\overrightarrow u = u\widehat i

$\therefore$

\left| {\overrightarrow v - \overrightarrow u } \right| = \sqrt {({u^2} - 2gL) + {u^2}}

= \sqrt {2u - 2gL}

$\therefore$

x = 2

Q13

A stone of mass m, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

A the same throughout the motion.

B minimum at the highest position of the circular path.

C minimum at the lowest position of the circular path.

D minimum when the rope is in the horizontal position.

Correct Answer

Option B

Solution

At any

\theta :T - mg\cos \theta = {{m{v^2}} \over R}

\Rightarrow T = mg\cos \theta + {{m{v^2}} \over R}

Since v is constant, $\Rightarrow$ T will be minimum when cos $\theta$ is minimum. $\Rightarrow$ $\theta$ = 180

^\circ

corresponds to Tminimum.

Q14

A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N. If the maximum speed with which the stone can revolve is

{K \over \pi }

rev./min. The value of K is : (Assume the string is massless and unstretchable)

A 400

B 300

C 600

D 800

Correct Answer

Option C

Solution

T = m{\omega ^2}r

\Rightarrow 80 = 0.1 \times {\left( {2\pi \times {K \over \pi } \times {1 \over {60}}} \right)^2} \times 2

\Rightarrow {{800} \over 2} = {{{K^2}} \over {900}}

\Rightarrow K = 30 \times 20 = 600

Q15

A body of mass 200g is tied to a spring of spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s. Then the ratio of extension in the spring to its natural length will be :

A 1 : 2

B 2 : 3

C 2 : 5

D 1 : 1

Correct Answer

Option B

Solution

$\because k x=m \omega^{2}(\ell+x)$ $12.5(x)=\dfrac{1}{5}(5)^{2}(\ell+x)$ $\Rightarrow \dfrac{5}{2} x=\ell+x$ $\Rightarrow \dfrac{3}{2} x=\ell$ $\Rightarrow \dfrac{x}{\ell}=\dfrac{2}{3}$

Q16

A coin placed on a rotating table just slips when it is placed at a distance of

1 \mathrm{~cm}

from the center. If the angular velocity of the table in halved, it will just slip when placed at a distance of _________ from the centre :

A 1 cm

B 8 cm

C 4 cm

D 2 cm

Correct Answer

Option C

Solution

When a coin is placed on a rotating table and is just about to slip, the centrifugal force acting on the coin equals the maximum static friction force.

Let's denote the mass of the coin as

m

, the initial angular velocity as

\omega_1

, and the final angular velocity as

\omega_2

. Initially, when the coin is placed at a distance of 1 cm from the center, the centrifugal force acting on the coin is:

F_1 = m r_1 \omega_1^2

where

r_1 = 1 \mathrm{~cm}

. When the angular velocity is halved (

\omega_2 = \frac{1}{2}\omega_1

), the centrifugal force acting on the coin when it just slips is:

F_2 = m r_2 \omega_2^2 = m r_2 \left(\frac{1}{2}\omega_1\right)^2

Since the coin is just about to slip in both cases, the maximum static friction force remains the same.

Therefore, we can equate the centrifugal forces:

m r_1 \omega_1^2 = m r_2 \left(\frac{1}{2}\omega_1\right)^2

Canceling the mass

m

and the initial angular velocity

\omega_1^2

from both sides, we get:

r_1 = r_2 \left(\frac{1}{2}\right)^2

Now, we can solve for

r_2

:

r_2 = \frac{r_1}{\left(\frac{1}{2}\right)^2} = \frac{1 \mathrm{~cm}}{\frac{1}{4}} = 4 \mathrm{~cm}

So, the coin will just slip when placed at a distance of 4 cm from the center when the angular velocity is halved.

Q17

A small block of mass

100 \mathrm{~g}

is tied to a spring of spring constant

7.5 \mathrm{~N} / \mathrm{m}

and length

20 \mathrm{~cm}

. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity

5 ~\mathrm{rad} / \mathrm{s}

about point

\mathrm{A}

, then tension in the spring is -

A 0.50 N

B 1.5 N

C 0.75 N

D 0.25 N

Correct Answer

Option C

Solution

In this problem, the spring is stretched due to the circular motion of the block, so the effective radius of the circular motion becomes the natural length of the spring plus the extension in the spring $(r=0.2+x)$ .

The spring force, which is also the centripetal force, is given by $F_c=Kx=m\omega^2 r$ , where $K$ is the spring constant, $x$ is the extension of the spring, $m$ is the mass of the block, $\omega$ is the angular velocity, and $r$ is the radius of the circular path.

Plugging in the values and solving for $x$ and $Kx$ gives the extension in the spring as $x = 0.1$ m and the tension in the spring (which is the spring force) as $Kx=7.5 \times 0.1 = 0.75$ N.

Substituting the given values: $7.5x = 0.1 \cdot (5^2) \cdot (0.2 + x),$ which simplifies to: $7.5x = 5 \cdot (x + 0.2)$ Solving for $x$ gives $x = 0.1$ m.

The tension in the spring is then $kx = 7.5 \cdot 0.1 = 0.75$ N.

So, the correct answer is $0.75$ N.

Q18

A child of mass

5 \mathrm{~kg}

is going round a merry-go-round that makes 1 rotation in

3.14 \mathrm{~s}

. The radius of the merry-go-round is

2 \mathrm{~m}

. The centrifugal force on the child will be

A 50 N

B 80 N

C 100 N

D 40 N

Correct Answer

Option D

Solution

To calculate the centrifugal force acting on the child, we need to find the angular velocity of the merry-go-round and then apply the formula for centrifugal force.

The merry-go-round makes 1 rotation in 3.14 seconds, so its angular velocity ( $\omega$ ) can be calculated as:

\omega = \frac{2\pi}{T}

, where

T

is the time period for one rotation.

\omega = \frac{2\pi}{3.14} = 2 \, \text{radians/s}

Now, the formula for centrifugal force (

F

) is:

F = m \cdot r \cdot \omega^2

, where

m

is the mass of the child,

r

is the radius of the merry-go-round, and $\omega$ is the angular velocity.

F = 5\,\text{kg} \cdot 2\,\text{m} \cdot (2\,\text{radians/s})^2 = 5\,\text{kg} \cdot 2\,\text{m} \cdot 4\,(\text{radians/s})^2

F = 40\,\text{N}

Therefore, the centrifugal force on the child is

40\,\text{N}

.

Q19

A clock has

75 \mathrm{~cm}, 60 \mathrm{~cm}

long second hand and minute hand respectively. In 30 minutes duration the tip of second hand will travel

x

distance more than the tip of minute hand. The value of

x

in meter is nearly (Take

\pi=3.14

) :

A 118.9

B 140.5

C 139.4

D 220.0

Correct Answer

Option C

Solution

To determine the distance traveled by the tips of the second hand and the minute hand, we need to calculate the circumference of the circles they make.

Let's start by calculating the distances traveled by both hands over a period of 30 minutes.

First, the circumference formula is given by:

C = 2\pi r

For the second hand: The length of the second hand is

75 \space \text{cm}

. The second hand completes one full revolution every 60 seconds, so in 1 minute, the second hand travels:

\text{Distance per minute} = 2\pi \times 75 \text{ cm}

In 30 minutes, the second hand will travel:

\text{Total distance traveled by second hand} = 30 \times 2\pi \times 75 \text{ cm}

Now, let's calculate it with

\pi = 3.14

:

\text{Total distance traveled by second hand} = 30 \times 2 \times 3.14 \times 75 \text{ cm}

\text{Total distance traveled by second hand} = 30 \times 471 \text{ cm} = 14130 \text{ cm}

For the minute hand: The length of the minute hand is

60 \text{ cm}

. The minute hand completes one full revolution every 60 minutes, so in 30 minutes, the minute hand travels:

\text{Total distance traveled by minute hand} = 0.5 \times 2\pi \times 60 \text{ cm}

Again, let's calculate it with

\pi = 3.14

:

\text{Total distance traveled by minute hand} = 0.5 \times 2 \times 3.14 \times 60 \text{ cm}

\text{Total distance traveled by minute hand} = 0.5 \times 376.8 \text{ cm} = 188.4 \text{ cm}

The difference in distance

x

is:

x = 14130 \text{ cm} - 1884 \text{ cm}

x = 13941.6 \text{ cm}

Convert this distance into meters:

x = 13941.6 \text{ cm} \times \frac{1 \text{ meter}}{100 \text{ cm}}

x \approx 139.4 \text{ meters}

Thus, the value of

x

in meters is nearly 139.4 meters, making the correct option: Option C

Q20

A ball of mass

0.5 \mathrm{~kg}

is attached to a string of length

50 \mathrm{~cm}

. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is

400 \mathrm{~N}

. The maximum possible value of angular velocity of the ball in

\mathrm{rad} / \mathrm{s}

is, :

A 1600

B 20

C 40

D 1000

Correct Answer

Option C

Solution

To find the maximum possible angular velocity (ω) of the ball, we need to consider the maximum tension the string can bear without breaking.

This tension provides the centripetal force needed to keep the ball in a circular path.

The centripetal force (Fc) required for circular motion is given by the formula: $F_{c} = m r \omega^2$ Where: m = mass of the ball (0.5 kg) r = radius of the circle (0.5 m, since 50 cm = 0.5 m) ω = angular velocity in rad/s The maximum tension the string can bear is also the maximum centripetal force (Fc,max) that can be provided by the string, which is 400 N.

Now we can set up the equation with the given values: $400 = 0.5 \times 0.5 \times \omega^2$ $\Rightarrow$ $400 = 0.25 \times \omega^2$ $\Rightarrow$ $\omega^2 = \dfrac{400}{0.25}$ $\Rightarrow$ $\omega^2 = 1600$ $\Rightarrow$ $\omega = \sqrt{1600}$ $\Rightarrow$ $\omega = 40 \, \mathrm{rad/s}$ Therefore, the maximum possible angular velocity of the ball is 40 rad/s, which corresponds to Option C.