SATHEE: Rotational Motion Question 83

Rotational Motion Question 83

Question: A mass m moves in a circle on a smooth horizontal plane with velocity $v_{0}$ at a radius $R_{0}$ . The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius $\frac{R_{0}}{2}$ . The final value of the kinetic energy is

Options:

A) $m v_{0}^{2}$

B) $\frac{1}{4} m v_{0}^{2}$

C) $2 m v_{0}^{2}$

D) $\frac{1}{2} m v_{0}^{2}$

Show Answer

Answer:

Correct Answer: C

Solution:

Conserving angular momentum

$L_{i} = L_{t}$

$\Rightarrow$ $m v_{0} R_{0} = m v^{'} (\frac{R_{0}}{2})$

$\Rightarrow$ $v^{'} = 2 v_{0}$

So, final kinetic energy of the particle is

$K_{f} = \frac{1}{2} m v {^{'}}^{2} = \frac{1}{2} m {(2 v_{0})}^{2}$

$= 4 \frac{1}{2} m v_{0}^{2} = 2 m v_{0}^{2}$

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Rotational Motion Question 83

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu