At time t=0 , a disk of radius 1 ~m starts to roll without slipping on a horizontal plane with an angular acceleration of \alpha=\frac23 rad \,s^-2 . A small stone is stuck to the disk. At t=0 , it is at the contact point of the disk and the plane. Later, at time t=\sqrt\pi \,s , the stone detaches itself and flies off tangentially from the disk. The maximum height (in m ) reached by the stone measured from the plane is \frac12+\fracx10 . The value of x is \underline\hspace2cm , [Take g=10 ~m ~s^-2 .]

Rotational Motion PYQ — JEE Advanced 2022 · Shift 1 · Q41 with Solution

View written solutionFree

Correct answer: 0.48TO0.56

Given data

Radius of disk: $R=1\,\text{m}$
Angular acceleration: $\alpha=\dfrac{2}{3}\,\text{rad s}^{-2}$
Initial angular velocity: since it "starts to roll", take $\omega_0=0$
Time of detachment: $t=\sqrt{\pi}\,\text{s}$
Rolling without slipping: $v_C=R\omega$
Gravity: $g=10\,\text{m s}^{-2}$

We need the maximum height from the plane reached by the stone after it detaches.

Angular position and angular speed at detachment

Since angular acceleration is constant,

\omega=\alpha t=\frac{2}{3}\sqrt{\pi}

Angular displacement from $t=0$ to $t=\sqrt{\pi}$ is

\theta=\frac{1}{2}\alpha t^2 =\frac{1}{2}\cdot \frac{2}{3}\cdot \pi =\frac{\pi}{3}

So the disk has rotated by $\dfrac{\pi}{3}$ clockwise.

Position of the stone at detachment

Initially the stone is at the contact point, i.e. at the bottom of the disk.

As the wheel rolls to the right, the radius joining center to the stone rotates clockwise by angle $\theta=\pi/3$ .

Take the center of the disk at detachment as origin for relative position. Then the stone relative to center is

\vec r = (-\sin\theta,\,-\cos\theta)

with $R=1$ .

Hence its vertical coordinate relative to center is

y_{rel}=-\cos\theta=-\cos\frac{\pi}{3}=-\frac{1}{2}

Since the center is always at height $R=1$ above the plane,

y_0=1-\frac{1}{2}=\frac{1}{2}

So at detachment the stone is at height

\boxed{\frac{1}{2}\,\text{m}}

above the plane.

Velocity of the stone at detachment

Velocity of the center:

\vec v_C=R\omega\,\hat i=\omega\hat i

The angular velocity is clockwise, so

\vec\omega=-\omega\hat k

Velocity of point relative to center:

\vec v_{rel}=\vec\omega\times\vec r

Using

\vec r=(-\sin\theta)\hat i+(-\cos\theta)\hat j

we get

\vec v_{rel}=(-\omega\hat k)\times\big[(-\sin\theta)\hat i+(-\cos\theta)\hat j\big]

which gives the vertical component

(v_{rel})_y=\omega\sin\theta

Since the center has no vertical velocity, the stone’s initial vertical velocity at detachment is

v_y=\omega\sin\theta

Now substitute values:

v_y=\frac{2}{3}\sqrt{\pi}\cdot \sin\frac{\pi}{3} =\frac{2}{3}\sqrt{\pi}\cdot \frac{\sqrt3}{2} =\frac{\sqrt{3\pi}}{3}

v_y^2=\frac{3\pi}{9}=\frac{\pi}{3}

Rise after detachment

Maximum additional rise in projectile motion is

\Delta h=\frac{v_y^2}{2g} =\frac{\pi/3}{20} =\frac{\pi}{60}

Therefore maximum height from the plane is

H=y_0+\Delta h =\frac{1}{2}+\frac{\pi}{60}

Given that

H=\frac{1}{2}+\frac{x}{10}

\frac{x}{10}=\frac{\pi}{60} \implies x=\frac{\pi}{6} approx 0.5236

Thus,

\boxed{x\approx 0.524}

Comparison with stored answer

Stored correct answer: $0.48$ to $0.56$

Our value $x\approx 0.524$ lies in this range, so it agrees.

More from Rotational Motion

View written solutionFree

Correct answer: 0.48TO0.56

Given data

Radius of disk: $R=1\,\text{m}$
Angular acceleration: $\alpha=\dfrac{2}{3}\,\text{rad s}^{-2}$
Initial angular velocity: since it "starts to roll", take $\omega_0=0$
Time of detachment: $t=\sqrt{\pi}\,\text{s}$
Rolling without slipping: $v_C=R\omega$
Gravity: $g=10\,\text{m s}^{-2}$

We need the maximum height from the plane reached by the stone after it detaches.

Angular position and angular speed at detachment

Since angular acceleration is constant,

\omega=\alpha t=\frac{2}{3}\sqrt{\pi}

Angular displacement from $t=0$ to $t=\sqrt{\pi}$ is

\theta=\frac{1}{2}\alpha t^2 =\frac{1}{2}\cdot \frac{2}{3}\cdot \pi =\frac{\pi}{3}

So the disk has rotated by $\dfrac{\pi}{3}$ clockwise.

Position of the stone at detachment

Initially the stone is at the contact point, i.e. at the bottom of the disk.

As the wheel rolls to the right, the radius joining center to the stone rotates clockwise by angle $\theta=\pi/3$ .

Take the center of the disk at detachment as origin for relative position. Then the stone relative to center is

\vec r = (-\sin\theta,\,-\cos\theta)

with $R=1$ .

Hence its vertical coordinate relative to center is

y_{rel}=-\cos\theta=-\cos\frac{\pi}{3}=-\frac{1}{2}

Since the center is always at height $R=1$ above the plane,

y_0=1-\frac{1}{2}=\frac{1}{2}

So at detachment the stone is at height

\boxed{\frac{1}{2}\,\text{m}}

above the plane.

Velocity of the stone at detachment

Velocity of the center:

\vec v_C=R\omega\,\hat i=\omega\hat i

The angular velocity is clockwise, so

\vec\omega=-\omega\hat k

Velocity of point relative to center:

\vec v_{rel}=\vec\omega\times\vec r

Using

\vec r=(-\sin\theta)\hat i+(-\cos\theta)\hat j

we get

\vec v_{rel}=(-\omega\hat k)\times\big[(-\sin\theta)\hat i+(-\cos\theta)\hat j\big]

which gives the vertical component

(v_{rel})_y=\omega\sin\theta

Since the center has no vertical velocity, the stone’s initial vertical velocity at detachment is

v_y=\omega\sin\theta

Now substitute values:

v_y=\frac{2}{3}\sqrt{\pi}\cdot \sin\frac{\pi}{3} =\frac{2}{3}\sqrt{\pi}\cdot \frac{\sqrt3}{2} =\frac{\sqrt{3\pi}}{3}

v_y^2=\frac{3\pi}{9}=\frac{\pi}{3}

Rise after detachment

Maximum additional rise in projectile motion is

\Delta h=\frac{v_y^2}{2g} =\frac{\pi/3}{20} =\frac{\pi}{60}

Therefore maximum height from the plane is

H=y_0+\Delta h =\frac{1}{2}+\frac{\pi}{60}

Given that

H=\frac{1}{2}+\frac{x}{10}

\frac{x}{10}=\frac{\pi}{60} \implies x=\frac{\pi}{6} approx 0.5236

Thus,

\boxed{x\approx 0.524}

Comparison with stored answer

Stored correct answer: $0.48$ to $0.56$

Our value $x\approx 0.524$ lies in this range, so it agrees.

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