Question

smic
32.
A ring rolls down from the top of an inclined plane of vertical height h. It will reach the ground after
time
1
4h
1
2h
1
3h
1)
2)
3)
1 14h
4)
sin Ꮎ Ꮙ5g
sin og
sin
00
sin 01
g​

Answer 1

Explanation:

The moment of inertia of ring, disc, solid sphere and hollow sphere are mr

2

,

2

1

mr

2

,

5

2

mr

2

and

3

2

mr

2

respectively. As the moment of inertia is the resistance to rotational motion we see that the object with least moment of inertia will reach the bottom first. Thus the order of reaching the bottom is S,D,HandR.

The gain in total kinetic energy is equal to loss in gravitational potential energy which in turn in same for all 4 objects and equal to mgh, where h is height of inclined plane.

Answer 2

The time taken by the ring to reach the ground will be option (A) $\frac{1}{sin\alpha } \sqrt{\frac{4h}{g} }$.

Given:

Vertical height of the slope = h

To find:

Time taken by the ring to reach the ground.

Explanation:

Step 1

We have been given that a ring rolls down an inclined plane of vertical height $'h'$ . The inclined plane makes an angle $\alpha$ with the horizontal.

Therefore,

Length of the inclined plane, or ;

The distance to be traveled by the ring will be  $l=\frac{h}{sin\alpha }$.

We know, an important formula for time taken by any object to reach the ground from an inclined plane that is

$T=\frac{\sqrt{2l(1+\frac{K^{2} }{R^{2} } )} }{g(sin\alpha) }$

Where,

$l=$ the distance traveled by the object

$K=$ Radius of gyration of the object

Step 2

We have,

Moment of inertia of the ring $=MR^{2}$

Therefore,

The radius of gyration of the ring will be

$MK^{2} =MR^{2}$

Hence, $K=R$

Substituting the known values in the equation, we get

$T=\sqrt{\frac{2(\frac{h}{sin\alpha } )(1+\frac{R^{2} }{R^{2} } )}{g(sin\alpha )}$

$T=\sqrt{\frac{\frac{2h}{sin\alpha }(2) }{g(sin\alpha )} }$

$T=\sqrt{\frac{4h}{gsin^{2}\alpha } }$   $;$ $or$

$T=\frac{1}{sin\alpha } \sqrt{\frac{4h}{g} }$

Final answer:

Hence, the time taken by the ring to reach the ground will be option (A) $\frac{1}{sin\alpha } \sqrt{\frac{4h}{g} }$.