Circular Motion — Page 3 | JEE Physics

Q21

A particle moving in a circle of radius

\mathrm{R}

with uniform speed takes time

\mathrm{T}

to complete one revolution. If this particle is projected with the same speed at an angle

\theta

to the horizontal, the maximum height attained by it is equal to

4 R

. The angle of projection

\theta

is then given by :

A

\sin ^{-1}\left[\dfrac{2 \mathrm{gT}^2}{\pi^2 \mathrm{R}}\right]^{\dfrac{1}{2}}

B

\sin ^{-1}\left[\dfrac{\pi^2 \mathrm{R}}{2 \mathrm{gT}^2}\right]^{\dfrac{1}{2}}

C

\cos ^{-1}\left[\dfrac{\pi \mathrm{R}}{2 \mathrm{gT}^2}\right]^{\dfrac{1}{2}}

D

\cos ^{-1}\left[\dfrac{2 \mathrm{gT}^2}{\pi^2 \mathrm{R}}\right]^{\dfrac{1}{2}}

Correct Answer

Option A

Solution

To solve for the angle of projection $\theta$ , we will first establish the relationship between the variables given and then derive the formula using kinematics.

The time $T$ for one revolution at speed $v$ in a circle of radius $R$ is related to the circumference of the circle by the formula:

v = \frac{2\pi R}{T}

When the particle is projected with the same speed $v$ at an angle $\theta$ to the horizontal, its vertical component of velocity is given by $v_y=v\sin\theta$ .

The maximum height $H$ reached by the projectile can be found from the kinematic equation:

H = \frac{v_y^2}{2g} = \frac{(v\sin\theta)^2}{2g}

We are given that the maximum height attained $H$ is equal to $4R$ , so:

4R = \frac{(v\sin\theta)^2}{2g}

Substitute $v$ from the first equation into the second one:

4R = \frac{((\frac{2\pi R}{T})\sin\theta)^2}{2g}

4R = \frac{(2\pi R\sin\theta)^2}{2gT^2}

4R = \frac{4\pi^2 R^2 \sin^2\theta}{2gT^2}

2gT^2 = \pi^2 R \sin^2\theta

Now solve for $\sin\theta$ :

\sin\theta =\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2}

Finally, to solve for $\theta$ , take the inverse sine of both sides:

\theta = \sin^{-1}\left(\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2}\right)

Therefore, the correct answer is: Option A

\sin^{-1}\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2}

Q22

A stone of mass

900 \mathrm{~g}

is tied to a string and moved in a vertical circle of radius

1 \mathrm{~m}

making

10 \mathrm{~rpm}

. The tension in the string, when the stone is at the lowest point is (if

\pi^2=9.8

and

g=9.8 \mathrm{~m} / \mathrm{s}^2

) :

A 17.8 N

B 97 N

C 9.8 N

D 8.82 N

Correct Answer

Option C

Solution

Given that

\begin{aligned} & \mathrm{m}=900 \mathrm{~gm}=\frac{900}{1000} \mathrm{~kg}=\frac{9}{10} \mathrm{~kg} \\ & \mathrm{r}=1 \mathrm{~m} \\ & \omega=\frac{2 \pi \mathrm{N}}{60}=\frac{2 \pi(10)}{60}=\frac{\pi}{3} \mathrm{rad} / \mathrm{sec} \\ & \mathrm{T}-\mathrm{mg}=\mathrm{mr} \omega^2 \\ & \mathrm{~T}=\mathrm{mg}+\mathrm{mr} \omega^2 \\ & =\frac{9}{10} \times 9.8+\frac{9}{10} \times 1\left(\frac{\pi}{3}\right)^2 \\ & =8.82+\frac{9}{10} \times \frac{\pi^2}{9} \\ & =8.82+0.98 \\ & =9.80 \mathrm{~N} \end{aligned}

Q23

A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius

9 \mathrm{~m}

and completes 120 resolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is (in

\mathrm{m} / \mathrm{s}^2

) :

A

4 \pi^2 \mathrm{~ms}^{-2}

B

16 \pi^2 \mathrm{~ms}^{-2}

C

57600 \pi^2 \mathrm{~ms}^{-2}

D Zero

Correct Answer

Option B

Solution

First, let's calculate the centripetal acceleration experienced by the monkey while the man does cycling smoothly on a circular track.

The formula for centripetal acceleration ( $a_c$ ) is given by: $a_c = \dfrac{v^2}{r}$ where $v$ is the velocity of the monkey and the man cycling, $r$ is the radius of the circular track.

To find $v$ , we first need to find the circumference of the circle, which is given by: $C = 2\pi r$ Given the radius $r = 9 \, \mathrm{m}$ , the circumference $C$ is: $C = 2\pi \times 9 = 18\pi \, \mathrm{m}$ Then, we determine the total distance travelled by calculating how many times they complete the circle in the given time.

With 120 revolutions in 3 minutes (180 seconds), the total distance $D$ travelled is: $D = 120 \times C = 120 \times 18\pi = 2160\pi \, \mathrm{m}$ To find the speed $v$ , which is distance over time, we divide the total distance by the total time in seconds: $v = \dfrac{D}{t} = \dfrac{2160\pi}{180} = 12\pi \, \mathrm{m/s}$ Now, using the formula for centripetal acceleration $a_c = \dfrac{v^2}{r}$ , we can plug in the values: $a_c = \dfrac{(12\pi)^2}{9} = \dfrac{144\pi^2}{9} = 16\pi^2 \mathrm{~m/s}^2$ Therefore, the magnitude of the centripetal acceleration of the monkey is $16\pi^2 \, \mathrm{m/s}^2$ , which corresponds to Option B.

Q24

A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is

8 \mathrm{~m} / \mathrm{s}

. The speed of the particle on the rim of the wheel at the same level as the centre of wheel, will be :

A

4 \sqrt{2} \mathrm{~m} / \mathrm{s}

B

8 \mathrm{~m} / \mathrm{s}

C

4 \mathrm{~m} / \mathrm{s}

D

8 \sqrt{2} \mathrm{~m} / \mathrm{s}

Correct Answer

Option A

Solution

If $V_B=2 V$ Point A is instantaneous center of rotation Given $V_B=8 \mathrm{~m} / \mathrm{s}$

\begin{aligned} & \mathrm{V}=4 \mathrm{~m} / \mathrm{s} \\ & \mathrm{~V}_{\mathrm{P}}=\sqrt{2} \mathrm{v} \Rightarrow V_{\mathrm{p}}=4 \sqrt{2} \mathrm{~m} / \mathrm{s} \\ & \operatorname{correct}(1) \end{aligned}

Q25

A car is moving with a constant speed of 20 m/s in a circular horizontal track of radius 40 m. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be : (Take g = 10 m/s

^2

)

A

\dfrac{\pi}{2}

B

\dfrac{\pi}{6}

C

\dfrac{\pi}{4}

D

\dfrac{\pi}{3}

Correct Answer

Option C

Solution

In car’s frame, FBD of bob where $a_{P}=$ Pseudoforce or centrifugal force $\theta=\tan ^{-1}\left(\dfrac{a_{P}}{g}\right)=\tan ^{-1}\left(\dfrac{v^{2}}{R g}\right)=\tan ^{-1}\left(\dfrac{400}{40 \times 10}\right)$ $=45^{\circ}$

Q26

A body is projected at t = 0 with a velocity 10 ms–1 at an angle of 60o with the horizontal. The radius of curvature of its trajectory at t = 1s is R. neglecting air resistance and taking acceleration due to gravity g = 10 ms–2, the value of R is :

A 2.8 m

B 5.1 m

C 2.5 m

D 10.3 m

Correct Answer

Option A

Solution

vx = 10cos60o = 5 m/s y = 10cos30o =

5\sqrt 3

m/s velocity after t = 1 sec. vx = 5 m/s vy =

\left| {\left( {5\sqrt 3 - 10} \right)} \right|

m/s = 10 $-$ 5

\sqrt 3

an =

{{{v^2}} \over R} \Rightarrow \,R\,

=

{{v_x^2 + v_y^2} \over {{a_n}}}

=

{{25 + 100 + 75 - 100\sqrt 3 } \over {10\cos \theta }}

tan $\theta$ =

{{10 - 5\sqrt 3 } \over 5}

= 2 $-$

{\sqrt 3 }

$\Rightarrow$ $\theta$ = 15o R =

{{100\left( {2 - \sqrt 3 } \right)} \over {10\cos 15}} = 2.8m

</b

Q27

For a particle in uniform circular motion, the acceleration

\overrightarrow a

at any point P(R,

\theta

) on the circular path of radius R is (when

\theta

is measured from the positive x-axis and v is uniform speed) :

A

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

B

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

As the particle in uniform circular motion experiences only centripetal acceleration of magnitude $\omega$ 2R or

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

directed towards centre so from diagram,

\overrightarrow a = {{{v^2}} \over R}\cos \theta ( - \widehat i) + {{{v^2}} \over R}\sin ( - \widehat j)

Q28

An object moves at a constant speed along a circular path in a horizontal plane with center at the origin. When the object is at

x=+2~\mathrm{m}

, its velocity is

\mathrm{ - 4\widehat j}

m/s. The object's velocity (v) and acceleration (a) at

x=-2~\mathrm{m}

will be

A

v=4\mathrm{\widehat i~m/s},a=8\mathrm{\widehat j~m/s^2}

B

v=4\mathrm{\widehat j~m/s},a=8\mathrm{\widehat i~m/s^2}

C

v=-4\mathrm{\widehat i~m/s},a=-8\mathrm{\widehat j~m/s^2}

D

v=-4\mathrm{\widehat j~m/s},a=8\mathrm{\widehat i~m/s^2}

Correct Answer

Option B

Solution

\overrightarrow v = 4\widehat j

(m/s)

a = {{{v^2}} \over R} = {{16} \over 2} = 8

m/s

^2

\overrightarrow a = 8\left( {m/{s^2}} \right)\left( {\widehat i} \right)

Q29

A clock has a continuously moving second's hand of 0.1 m length. The average acceleration of the tip of the hand (in units of ms–2) is of the order of :

A 10-3

B 10-1

C 10-2

D 10-4

Correct Answer

Option A

Solution

R = 0.1 m $\omega$ =

{{2\pi } \over T}

=

{{2\pi } \over {60}}

= 0.105 rad/sec a =

{\omega ^2}R

= (0.105)2(0.1) = 0.0011 = 1.1 $\times$ 10-3 Average acceleration is of the order of 10–3.

Q30

A body is moving with constant speed, in a circle of radius

10 \mathrm{~m}

. The body completes one revolution in

4 \mathrm{~s}

. At the end of 3rd second, the displacement of body (in

\mathrm{m}

) from its starting point is :

A

15 \pi

B 30

C

10 \sqrt{2}

D

5 \pi

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{\omega}=\frac{2 \pi}{\mathrm{T}}=\frac{2 \pi}{4}=\frac{\pi}{2} \mathrm{rad} / \mathrm{s} \\\\ & \theta=\mathrm{\omega t} \\\\ & \theta=\frac{\pi}{2} \times 3 \\\\ & \theta=\frac{3 \pi}{2} \mathrm{rad} \end{aligned}

\begin{aligned} & r=10 \mathrm{~m} \\\\ & T=4 \mathrm{sec} \\\\ & d=\sqrt{2}(10) \mathrm{m} \end{aligned}