Question 453.
The moment of inertia of a hollow sphere of mass m and radius r about an axis through its centre is (2mr^2)/3. A hollow sphere is bowled across a horizontal surface. The sphere initially slides with a velocity of u. After a short interval, it starts to roll without slipping.
Show that the velocity of the ball when it starts to roll is √(3/5) u.
Answer 453.
Kinetic energy of the ball when it slides without rolling
= (1/2) mu^2
When it starts rolling without slipping, it has both linear as well as rotational motion and the above kinetic energy gets converted into kinetic energy of linear and rotational motion
If v = velocity when it rolls without slipping, then
(1/2) mu^2 = (1/2)mv^2 + (1/2) I ω^2
=> mu^2 = mv^2 + (2/3) mr^2ω^2
=> u^2 = v^2 + (2/3) v^2 ... [because ωr = v]
=> u^2 = 5/3v^2
=> v^2 = 3/5u^2
=> v = √(3/5) u.
Link to YA!
The moment of inertia of a hollow sphere of mass m and radius r about an axis through its centre is (2mr^2)/3. A hollow sphere is bowled across a horizontal surface. The sphere initially slides with a velocity of u. After a short interval, it starts to roll without slipping.
Show that the velocity of the ball when it starts to roll is √(3/5) u.
Answer 453.
Kinetic energy of the ball when it slides without rolling
= (1/2) mu^2
When it starts rolling without slipping, it has both linear as well as rotational motion and the above kinetic energy gets converted into kinetic energy of linear and rotational motion
If v = velocity when it rolls without slipping, then
(1/2) mu^2 = (1/2)mv^2 + (1/2) I ω^2
=> mu^2 = mv^2 + (2/3) mr^2ω^2
=> u^2 = v^2 + (2/3) v^2 ... [because ωr = v]
=> u^2 = 5/3v^2
=> v^2 = 3/5u^2
=> v = √(3/5) u.
Link to YA!
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