Advertisements
Advertisements
Question
Obtain an expression for total kinetic energy of a rolling body in the form
`1/2 (MV^2)[1+K^2/R^2]`
Solution
Let M and R be the mass and radius of the body, V is the translation speed, ω is the angular speed and I is the moment of inertia of the body about an axis passing through the centre of mass.
Kinetic energy of rotation, `E_R=1/2MV^2`
Kinetic energy of translation, `E_r=1/2Iomega^2`
Thus, the total kinetic energy 'E' of the rolling body is
E = ER + Er
`=1/2MV^2+1/2Iomega^2`
`=1/2MV^2+1/2MK^2omega^2 ......(I=MK^2 " and K is the radius of gyration")`
`=1/2MR^2omega^2+1/2MK^2omega^2 ........(V=Romega)`
`:.E=1/2Momega^2(R^2+K^2)`
`:.E=1/2MV^2/R^2(R^2+K^2)`
`E=1/2MV^2(1+K^2/R^2)`
Hence proved.
APPEARS IN
RELATED QUESTIONS
The kinetic energy of a rotating body depends upon................
- distribution of mass only.
- angular speed only.
- distribution of mass and angular speed.
- angular acceleration only.
The body is rotating with uniform angular velocity (w) having rotational kinetic energy (E). Its angular momentum (L) is: ...............
a) `(2E)/ω`
b) `E^2/ω`
c) `E/ω^2`
d) `E/(2ω)`
The kinetic energy of emitted photoelectorns is independent of ............
(a) frequency of incident radiation.
(b) intensity of incident radiation.
(c) wavelength of incident radiation
(d) collector plate potential
If ‘L’ is the angular momentum and ‘I’ is the moment of inertia of a rotating body, then `L^2/(2I)`represents its _____
(A) rotational P.E.
(B) total energy
(C) rotational K.E.
(D) translational K.E
Define radius of gyration. Write its physical significance.
The radius of gyration of a body about an axis, at a distance of 0.4 m from its centre of mass is 0.5 m. Find its radius of gyration about a parallel axis passing through its centre of mass.
