Home
Class 11
PHYSICS
The kinetic energy K of a rotating body ...

The kinetic energy K of a rotating body depends on its moment of inertia I and its angular speed `omega`. Assuming the relation to be `K=kI^(alpha) omega^b` where k is a dimensionless constatnt, find a and b. Moment of inertia of a spere about its diameter is `2/5Mr^2`.

Text Solution

Verified by Experts

The correct Answer is:
A, B
`K=KI^aomega^b
where K=` kinetic energy of rotating body and kk= dimensionless constant Dimensions of left side are `K=[ML^2T^-2]`Dimensions of right side are `I^a=[ML^2}^a, omega^b=[T^-1]^b`
According to principle of homogeneity of dimension
`[ML^2T^-2]=[ML^2][T^-1]^b` Equatin the dimensions of both sides wer get `2=2a and -2=-b
alpha =1 and b=2`
Promotional Banner

Similar Questions

Explore conceptually related problems

Percentage error in measuring the radius and mass of a solid sphere are 2 % & 1 % respectively. The error in measurement of moment of inertia about to its diameter is :-

Prove that moment of inertia of uniform ring of mass M and radius R about its geometric axis is MR^(2)

(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR^(2)//5 , where M is the mass of the sphere and R is the radius of the sphere. (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR^(2)//4 , find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2(MR^(2))/(5) , where M is the mass of the sphere and R is the radius of the sphere. (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be (MR^(2))/(4) , find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Two bodies have their moment of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momentum will be in the ratio :

If I_(1) is the moment of inertia of a uniform rod about an axis perpendicular to its length and passing through its one end. Now ring formed by bending the rod, if the moment of inertia about the diameter of ring I_(1) then (I_(1))/(I_(2)) = ……….

(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

A wheel having moment of inertia 2kgm^(2) about its vertical axis, rotates at the rate of 60 rpm about the axis. The torque which can stop the wheel.s rotation in one minute would be……………

The maximum percentage errors in the measurement of mass (M), radius (R) and angular velocity (omega) of a ring are 2%,1% and 1% respectively, then find the maximum percentage error in the measurement of its- (a) Moment of inertia (1=(1)/(2)MR^(2)) about its geometric axis. (b) Rotational kinetic energy (K=(1)/(2)Iomega^(2)) and (c) Angular momentum (J=I omega) about geometrical axis.

A circular disc of moment of inertia I_(1) is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed omega_(1) . Another disc of moment of inertia I_(2) is placed coaxially on the rotating disc. Initially the second disc has zero angular speed. Eventually both the discs rotate with a constant angular speed omega_(2) . The energy lost by the initially rotating disc to friction is ............