Question

A disc rolls down without slipping. From rest on 37 inclined.its linear acceleration is ​

Answer 1

Answer: g sin 37

Explanation: a=gsin37 = g(4/5)

Answer 2

The linear acceleration of the disc is 2g/5

Explanation:

Given:

A disc rolls down without slipping from rest on 37° inclined plane

To find out:

The linear acceleration of the disc

Solution:

We know that if a round body rolls down an inclined plane (angle of inclination θ) starting from rest, the linear acceleration is given by

$\boxed{a=\frac{g\sin\theta}{1+(k^2/R^2)}}$

Where k is the radius of gyration

R is the radius of the round body

We know that if mass of the disc is M and radius R then the moment of inertia of disc is

$I=\frac{1}{2}MR^2$

or, $Mk^2=\frac{1}{2}MR^2$

$\implies \frac{k^2}{R^2}=\frac{1}{2}$

Thus, the linear acceleration of the disc

$a=\frac{g\sin 37^\circ}{1+1/2}$

$\implies a=\frac{g\times (3/5)}{3/2}$

$\implies a=\frac{2g}{5}$

Hope this answer is helpful.

Know More:

Q: Thin uniform, circular ring is rolling down an inclined plane of inclination 30° without slipping. Its linear acceleration along the inclined plane will be:

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