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Accleration at different point on disc

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A disc of mass m and radius R is rolling without slipping on a horizontal fixed rough surface with tramlational velocity v as shown in the figure .

According to me KE should same wrt any point thai is

$(1/2)mv_{cm} ^2 + (1/2)I\omega ^2 = (3/4)mv^2$

In my book this is the KE given wrt point O and wrt point P it is given $(3/8)mv^2$

But how can be this possible ?

asked May 12, 2018 in Physics Problems by koolman (4,286 points)

1 Answer

1 vote
Best answer

The KE can be different in different frames of reference.

With respect to the stationary frame Oxy the KE is $\frac34 mv^2$ as per your calculation.

However, points C and P are moving relative to the frame Oxy. For example, C is moving with velocity v in the +x direction, so in the frame Cxy in which C is stationary the disk has no translational KE, only rotational KE. The total KE in this frame is $\frac14 mv^2$. The angular velocity $\omega$, and therefore also the rotational KE, are the same in all non-rotating reference frames.

The instantaneous velocity of point P in the frame Oxy is $\frac12 v$ in the +x direction, because P is half way between C (which has velocity $v$ in the +x direction) and the point of contact with the ground (which has velocity $0$). In the inertial frame Pxy which has the instantaneous velocity of P, the disk has velocity $\frac12 v$ and translational KE of $\frac12 m(\frac12 v)^2=\frac18 mv^2$. It still has rotational KE of $\frac14 mv^2=\frac28 mv^2$ so the total KE in this frame is $\frac38 mv^2$.

answered May 12, 2018 by sammy gerbil (28,466 points)
edited May 14, 2018 by sammy gerbil
Will the rotational KE will be same at every point on the disc ?
Yes. See my sentence in brackets at the end of the 3rd paragraph. Rotational KE is the same in all inertial frames of reference.