MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Angular Analogue of Linear Momentum

25.Show that ...

Question

25.Show that or derive an expression for the kinetic energy of a rotating body about a given axis is equal to 1/2 Lw, whre, L is angular momentum and w is angular velocity. i.e. K.E.=1/2 Lw

Open in App

Solution

Suggest Corrections

thumbs-up

24

Similar questions

Q. Derive an expression for the kinetic energy of a body of mass

M

rotating uniformly about a given axis. Hence show that rotational kinetic energy is

= \frac{1}{2 M} \times {(\frac{L}{K})}^{2}

Q. The moment of inertia of a body about a given axis is

1.2 k g m^{2}

. Initially, the body is at rest. In order to produce a rotational kinetic energy of

1500 J

, and an angular acceleration of

25 r a d s^{- 2}

, the torque that must be applied about that axis for a duration of:

Q. A body having a moment of inertia about its axis of rotation equal to 3

k g - m^{2}

is rotating with angular velocity of

3 r a d s^{- 1}

. Kinetic energy of this rotating body is same as that of a body of mass

27 k g

moving with velocity

v

v

Q. The moment of inertia of a body about a given axis is 1.2 kg

m^{2}

. Initially the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25

r a d / s^{2}

must be applied about that axis for a duration of

Q. The moment of inertia of a body about a given axis is

1.2

m^{2}

. To produce a rotational kinetic energy of

1500

J an angular acceleration of

25

s^{2}

must be applied for.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Intuition of Angular Momentum

PHYSICS

Watch in App

explore_more_icon

Explore more

Angular Analogue of Linear Momentum

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon