Question

Give me jee advance level question on circular motion

Answer 1

(a) $\theta = \arccos\left(\frac{2}{3}\right)$ (b) $v = \sqrt{\frac{2gR}{3}}$ (c) $T = \frac{26mg}{27}$ (d) $v_{bottom,min} = \sqrt{5gR}$

Answer 2

Here's a JEE Advanced level question on circular motion:

Question:

A small block of mass $m$ is released from rest at the top of a large fixed smooth sphere of radius $R$. It slides down the sphere. At the instant it leaves the surface of the sphere, it is attached to a light inextensible string of length $R$ (same as sphere radius) whose other end is fixed at the center of the sphere.

(a) Find the angle $\theta$ (with the vertical) at which the block leaves the surface of the sphere. (b) Find the speed of the block just as it leaves the surface of the sphere. (c) What is the tension in the string just after it becomes taut? Assume the string becomes taut instantaneously without energy loss, and the velocity component perpendicular to the string is conserved. (d) What is the minimum speed the block must have at the lowest point of its subsequent circular path (after the string becomes taut) to complete a full vertical circle?

Solution:

Let the center of the sphere be the origin. The block is released from rest at the top, so its initial height is $R$.

Part (a) and (b): Block leaving the sphere

1. Energy Conservation:
   Let the block be at an angle $\theta$ with the vertical. Its height above the center is $R\cos\theta$. The vertical distance fallen is $R - R\cos\theta$.
   By conservation of mechanical energy (since the sphere is smooth):
   Initial Energy (at top) = Final Energy (at angle $\theta$)
   $mgR + 0 = mgR\cos\theta + \frac{1}{2}mv^2$
   $\frac{1}{2}mv^2 = mgR(1 - \cos\theta)$
   $v^2 = 2gR(1 - \cos\theta)$
   (Equation 1)

2. Condition for leaving the surface:
   The block leaves the surface when the normal force $N$ exerted by the sphere becomes zero. At this point, the gravitational force component along the radius provides the necessary centripetal force.
   Applying Newton's second law in the radial direction (towards the center):
   $mg\cos\theta - N = \frac{mv^2}{R}$
   When the block leaves the surface, $N=0$:
   $mg\cos\theta = \frac{mv^2}{R}$
   (Equation 2)

3. Solving for $\theta$ and $v$:
   Substitute $v^2$ from Equation 1 into Equation 2:
   $mg\cos\theta = \frac{m}{R} [2gR(1 - \cos\theta)]$
   $g\cos\theta = 2g(1 - \cos\theta)$
   $\cos\theta = 2 - 2\cos\theta$
   $3\cos\theta = 2$
   $\cos\theta = \frac{2}{3}$

So, the angle at which the block leaves the sphere is $\theta = \arccos\left(\frac{2}{3}\right)$.

Now, substitute $\cos\theta = \frac{2}{3}$ back into Equation 1 to find the speed $v$:
$v^2 = 2gR\left(1 - \frac{2}{3}\right)$
$v^2 = 2gR\left(\frac{1}{3}\right)$
$v = \sqrt{\frac{2gR}{3}}$

Part (c): Tension in the string just after it becomes taut

1. Velocity components at leaving point:
   At the instant the block leaves the sphere, its velocity $\vec{v}$ makes an angle $\theta$ with the vertical.
   The string is of length $R$ and is attached to the center of the sphere. This means the block is now constrained to move in a vertical circle of radius $R$.
   The velocity component along the radius (towards the center) is $v_r = v \sin\theta$. This component is instantaneously lost when the string becomes taut (due to the inelastic constraint).
   The velocity component perpendicular to the string (tangential to the new circular path) is $v_t = v \cos\theta$. This component is conserved.
   So, the speed of the block just after the string becomes taut is $v' = v \cos\theta$.

2. Calculate $v'$:
   We have $v = \sqrt{\frac{2gR}{3}}$ and $\cos\theta = \frac{2}{3}$.
   $v' = \sqrt{\frac{2gR}{3}} \times \frac{2}{3} = \frac{2}{3}\sqrt{\frac{2gR}{3}}$

3. Newton's Second Law for Circular Motion:
   Just after the string becomes taut, the block is moving in a circle of radius $R$ at an angle $\theta$ with the vertical. The forces acting on it are tension $T$ (inwards along the string) and gravity $mg$ (vertically downwards).
   The component of gravity along the radius (outwards from the center) is $mg\cos\theta$.
   Applying Newton's second law in the radial direction (inwards is positive):
   $T - mg\cos\theta = \frac{mv'^2}{R}$
   $T = \frac{mv'^2}{R} + mg\cos\theta$

4. Calculate Tension $T$:
   Substitute $v'$ and $\cos\theta$:
   $T = \frac{m}{R} \left( \frac{2}{3}\sqrt{\frac{2gR}{3}} \right)^2 + mg\left(\frac{2}{3}\right)$
   $T = \frac{m}{R} \left( \frac{4}{9} \times \frac{2gR}{3} \right) + \frac{2mg}{3}$
   $T = \frac{8mgR}{27R} + \frac{2mg}{3}$
   $T = \frac{8mg}{27} + \frac{18mg}{27}$
   $T = \frac{26mg}{27}$

Part (d): Minimum speed at lowest point for full vertical circle

1. Condition for completing a vertical circle with a string:
   For a block attached to a string to complete a full vertical circle, the tension in the string must remain non-negative throughout the motion. The critical point is the topmost point, where the tension is minimum.
   Let $v_{top}$ be the speed at the topmost point. At this point, both tension $T_{top}$ and gravity $mg$ act downwards.
   $T_{top} + mg = \frac{mv_{top}^2}{R}$
   For the string not to slack, $T_{top} \ge 0$.
   Therefore, $\frac{mv_{top}^2}{R} \ge mg \implies v_{top}^2 \ge gR$.
   The minimum speed at the top is $v_{top,min} = \sqrt{gR}$.

2. Energy Conservation between lowest and topmost points:
   Let $v_{bottom}$ be the speed at the lowest point. We can use conservation of mechanical energy between the lowest point and the topmost point.
   Let the lowest point be the reference for potential energy ($PE=0$).
   At the lowest point: $PE_{bottom} = 0$, $KE_{bottom} = \frac{1}{2}mv_{bottom}^2$.
   At the topmost point: $PE_{top} = mg(2R)$ (since height difference is $2R$), $KE_{top} = \frac{1}{2}mv_{top,min}^2 = \frac{1}{2}m(gR)$.

$PE_{bottom} + KE_{bottom} = PE_{top} + KE_{top}$
$0 + \frac{1}{2}mv_{bottom}^2 = mg(2R) + \frac{1}{2}mgR$
$\frac{1}{2}mv_{bottom}^2 = \frac{5}{2}mgR$
$v_{bottom}^2 = 5gR$
$v_{bottom,min} = \sqrt{5gR}$

This is the minimum speed the block must have at the lowest point to complete a full vertical circle.

Summary of Answers:

(a) The angle $\theta$ at which the block leaves the surface of the sphere is $\theta = \arccos\left(\frac{2}{3}\right)$. (b) The speed of the block just as it leaves the surface of the sphere is $v = \sqrt{\frac{2gR}{3}}$. (c) The tension in the string just after it becomes taut is $T = \frac{26mg}{27}$. (d) The minimum speed the block must have at the lowest point of its subsequent circular path to complete a full vertical circle is $v_{bottom,min} = \sqrt{5gR}$.

Explanation of the solution:

1. Part (a) & (b): Apply energy conservation from the top of the sphere to an arbitrary angle $\theta$ to relate speed $v$ to $\theta$. Then, apply Newton's second law in the radial direction. The condition for leaving the surface is that the normal force becomes zero. Solve the two equations simultaneously for $\theta$ and $v$.
2. Part (c): When the string becomes taut, the velocity component perpendicular to the string (tangential to the new circular path) is conserved, while the radial component is lost. Calculate this conserved tangential velocity. Then, apply Newton's second law in the radial direction for the new circular motion, considering tension and the radial component of gravity.
3. Part (d): For a particle attached to a string to complete a vertical circle, the tension must remain non-negative. This critical condition occurs at the topmost point where the minimum speed is $\sqrt{gR}$. Use energy conservation between the lowest and topmost points to find the corresponding minimum speed at the lowest point.