Question

a disc of radius 2m and mass 100 kg rolls on a horizontal floor its centre of mass has speed of 20cm/s .how much work is needed to stop it​

Answer 1

Answer: 3 joule

Explanation: The work done to stop the disc = rotational K.E + translational K.E .

Rotational K.E = 1/2 I W^2

( Here , I = 1/2 m r^2 and W = V/r)

Rotational KE = 1 J

Translational KE =1/2 mv^2

= 2 J

So , total energy = 3 j , which is the work needed to stop it .

Answer 2

Given:

Radius, r = 2 m

Mass, m = 100 kg

Speed, s = 20cm/s

To Find:

$The \ work \ done \ to \ stop \ the \ disc = rotational \ K.E + translational \ K.E$

Here,

$L = \frac{1}{2} m r^2$

and,

$W = \frac{V}{r}$

Rotational K.E = $\frac{1}{2} LW^2$

⇒                      = $1 \ Joule$

Translational KE = $\frac{1}{2} mv^2$

⇒                          = $2 \ Joule$

Hence,

The total work done will be = $1+2$

⇒                                             = $3 \ Joule$

So that the correct answer will be "3 Joule".