Home
Class 11
PHYSICS
A child of mass m is standing, on the pe...

A child of mass `m` is standing, on the periphery of a circular platform of radius `R`, which can rotate about its central axis. The moment of inertia of the platform is `I`. Child jumps off from the platform with a velocity `u` tangentially relative to the platform. Find the angular speed of the platform after the child jumps off.

Text Solution

AI Generated Solution

To solve the problem, we will use the principle of conservation of angular momentum. Let's break down the solution step by step. ### Step 1: Understand the Initial Conditions Initially, the child and the platform are at rest. Therefore, the initial angular momentum of the system is zero. **Hint:** Remember that angular momentum is conserved when no external torques act on the system. ### Step 2: Define the Final Conditions ...
Promotional Banner

Topper's Solved these Questions

  • RIGID BODY DYNAMICS 2

    CENGAGE PHYSICS ENGLISH|Exercise Exercise 3.3|21 Videos

  • RIGID BODY DYNAMICS 2

    CENGAGE PHYSICS ENGLISH|Exercise Exercise 3.4|13 Videos

  • RIGID BODY DYNAMICS 2

    CENGAGE PHYSICS ENGLISH|Exercise Exercise 3.1|11 Videos

  • RIGID BODY DYNAMICS 1

    CENGAGE PHYSICS ENGLISH|Exercise Integer|11 Videos

  • SOUND WAVES AND DOPPLER EFFECT

    CENGAGE PHYSICS ENGLISH|Exercise Integer|16 Videos

Similar Questions

Explore conceptually related problems

A kid of mass M stands at the edge of a platform of radius R which can be freely rotated about its axis. The moment of inertia of the platform is I . The system is at rest when a friend throws as ball of mass m and the kid catches it. If the velocity of the ball is v horizontally along the tangent to the edge of the platform when it was caught by the kid find the angular speed of the platform after the event.

A man of mass m is standing on a platform of mass M kept on smooth ice. If the man starts moving on the platform with a speed v relative to the platform with what velocity relative to the ice does the platform recoil?

A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm). The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm) of the platform when the person reaches its centre___________.

A child with mass m is standing at the edge of a merry go round having moment of inertia I , radius R and initial angular velocity omega as shown in the figure. The child jumps off the edge of the merry go round with tangential velocity v with respect to the ground. The new angular velocity of the merry go round is

A flat car of mass M is at rest on a frictionless floor with a child of mass m standing at its edge. If the child jumps off from the car towards right with an initial velocity u , with respect to the car, find the velocity of the car after its jump.

A concentric hole of radius R/2 is cut from a thin circular plate of mass M and radius R. The moment of inertia of the remaining plate about its axis will be

A concentric hole of radius R/2 is cut from a thin circular plate of mass M and radius R. The moment of inertia of the remaining plate about its axis will be

Suppose in the previous problem, the child stays at rest on the platform and one of his friend throws a ball of mas m_(1) towards him with a velocity of v from a direction tangential to the platform and the child on the platform catches the ball. Find the angular velocity of the platform after he catches the ball.

A person is standing on the edge of a circular platform, which is moving with constant angular speed about an axis passing through its centre and perpendicular to the plane of platform. If person is moving along any radius towards axis of rotation then the angular velocity will:

A child is standing with folded hands at the center of a platform rotating about its central axis. The kinetic energy of the system is K . The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is: a) K/4 b) K/2 c) 2K d) 4K