Question

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

Answer 1

acceleration on an inclined plane 
a= g sinθ/ (1 + I/MR²)

for circular rings: I= MR²

so by putting the value in above equation we get
a= g sinθ/ 2

a = g sin30/2
a= g/4

Answer 2

The linear acceleration is $\frac{g}{4}$

Given:

Angle of inclination = 30 degree

Solution:

The linear acceleration along the inclined must be in order to let not the ring slip.

This acceleration on the inclined plane is given by the formula below:

$a=\frac{g \sin \theta}{1+\frac{I}{M R^{2}}}$

Since, I for circular rings is given below:

$I=M R^{2}$

Thereby, we get,

$a=\frac{g \sin \theta}{1+\frac{M R^{2}}{M R^{2}}}$

$a=\frac{g \sin \theta}{1+1}$

$a=\frac{g \sin \theta}{2}$

$\Rightarrow a=\frac{g \sin 30}{2}$

$a=0.25(g)$

$\therefore a=\frac{g}{4}$