NEET2019Physics-Rotational Motion
NEET 2019 Physics Rolling Motion MCQ Question
Type: MCQ-numerical-Medium-Class 11
A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it?
A
2 J
B
1 J
C
3 J
D
30 kJ
Correct Answer
Option C
Detailed Explanation
The total kinetic energy of the rolling disc is the sum of translational and rotational kinetic energy: KE = 1/2 mv² + 1/2 Iω². For a disc, I = 1/2 mr² and ω = v/r. Substituting the values, KE = 1 J.
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