Question 633: Centre of Mass and Rotational Motion

Question

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?

Accepted Answer

$$\omega  = {V \over R} = {{0.2} \over 2} = $$ 0.1 rad/sec
Apply the law of conservation of energy,
Work required = change in kinetic energy
Since, final KE = 0
And, initial KE = $${1 \over 2}m{v^2} + {1 \over 2}I{\omega ^2}$$
= $${1 \over 2}m{v^2} + {1 \over 2}{{M{l^2}} \over 2}{\omega ^2}$$
= $${1 \over 2}\left( {100} ight){\left( {0.2} ight)^2} + {1 \over 2}{{100 	imes {{\left( 2 ight)}^2}} \over 2}{\left( {0.1} ight)^2}$$
= 1 + $${1 \over 2} 	imes 4$$ = 3 J

Answer

1 J

Answer

2 J

Answer

3 J

Answer

30 kJ

Centre of Mass and Rotational Motion

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?

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