Question 459: Centre of Mass and Rotational Motion

Question

A circular disc reaches from top to bottom of an inclined plane of length 'L'. When it slips down the plane, it makes time 't1'. When it rolls down the plane, it takes time t2. The value of $${{{t_2}} \over {{t_1}}}$$ is $$\sqrt {{3 \over x}} $$. The value of x will be _______________.

Accepted Answer

According to question, a circular disc reaches from top to bottom of an inclined plane of length L. This can be shown as
When the disc slips down the inclined plane, it takes time t
1
. Therefore, in this case its acceleration, a
1
= g sin$$	heta$$
$$\because$$ s = ut
1
+ $${1 \over 2}$$a
1
t$$_1^2$$
$$\Rightarrow$$ s = $${1 \over 2}$$ g sin$$	heta$$t$$_1^2$$ .... (i)
And when the disc rolls down the inclined plane, it takes time t
2
. Therefore in this case, its acceleration,
$${a_2} = {{g\sin 	heta } \over {1 + {{{K^2}} \over {{R^2}}}}} = {{g\sin 	heta } \over {1 + {1 \over 2}}} = {2 \over 3}g\sin 	heta $$ [$$\because$$ for disc, $${{{K^2}} \over {{R^2}}} = {1 \over 2}$$]
$$\because$$ s = ut
2
+ $${1 \over 2}$$ a
2
t$$_2^2$$ = $${1 \over 2}$$ . $${2 \over 3}$$ g sin$$	heta$$ t$$_2^2$$ .... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$${{{t_2}} \over {{t_1}}} = \sqrt {{3 \over 2}} $$ .... (iii)
According to question, value of $${{{t_2}} \over {{t_1}}}$$ is $$\sqrt {{3 \over x}} $$.
Comparing it with Eq. (iii), we get x = 2

Centre of Mass and Rotational Motion

A circular disc reaches from top to bottom of an inclined plane of length 'L'. When it slips down the plane, it makes time 't1'. When it rolls down the plane, it takes time t2. The value of ${{{t_2}} \over {{t_1}}} is \sqrt {{3 \over x}} $. The value of x will be _______________.

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