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Starting from rest, an object rolls down along an incline that rises by 3 in every 5 (along with it). The object gains a speed of sqrt10 m/s as it travels a distance of 5/3 m along the incline.

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Question

Starting from rest, an object rolls down along an incline that rises by 3 in every 5 (along with it). The object gains a speed of `sqrt 10` m/s as it travels a distance of `5/3` m along the incline. What can be the possible shape/s of the object? 

Numerical
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Solution

Given: Incline that rises by 3 in every 5 is sin θ = `3/5`

The object gains a speed of v = `sqrt10` m/s

It travels a distance along the incline s = `5/3` m

To find: The shape of the possible object, i.e., to find out the ratio of `K^2/R^2` which will determine the possible rolling object.

Solution: We have, `sin theta = 3/5`

And Linear distance travelled along the plane is s = `h/sin theta`

Hence,

h = s sin θ = `5/3 xx 3/5` = 1

The velocity of rolling body is given by,

v = `sqrt((2gh)/(1 + K^2/R^2))`

Comparing we get,

`sqrt 10 = sqrt((2gh)/(1 + K^2/R^2))`

10 = `(2 xx 10 xx 1)/(1 + (K^2)/R^2)`

`(1 + (K^2)/R^2)` = 2

`(K^2)/R^2` = 1

Hence object must be Ring or hollow cylinder.

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Rotational Dynamics
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Chapter 1: Rotational Dynamics - Exercises [Page 25]

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Balbharati Physics [English] Standard 12 Maharashtra State Board
Chapter 1 Rotational Dynamics
Exercises | Q 22 | Page 25

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