Question

A disc of mass 2kg is rolling on a horizontal surface without slipping with a velocity of 0.1m/s. what is its rotational kinetic energy

Answer 1

Answer:

0.005J

Explanation:

Rotational KE =(1/2)*I*ω^2

I=(1/2)*M*R^2

ω=v/R

v=0.1m/s

M=2kg

substituting ω in KE eqn;

RKE=(1/4)*M*v^2

RKE=0.005J

Answer 2

Given :

Mass of disc = 2kg

Velocity of disc = 0.1m/s

To Find :

Rotational kinetic energy of the disc$.$

Solution :

■ Rotational kinetic energy of object is given by

• k = 1/2 Iω² .......... (I)

✵ ω denotes angular velocity

✵ I denotes moment of inertia

■ Moment of inertia of a circular disc of radius r rotating about an axis passing through its centre and perpendicular to the plane is given by

• I = MR²/2

✵ M denotes mass of disc

✵ R denotes radius of disc

■ Relation b/w linear velocity and angular velocity is given by

• ω = v/R

By substituting all values in equation (I),

➠ k = 1/2 (MR²/2) (v/R)²

➠ k = 1/4 × Mv²

➠ k = 1/4 × (2)(0.1)²

➠ k = 0.005 J