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 't
1
'. When it rolls down the plane, it takes time t
2
. 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 't 1 '. When it rolls down the plane, it takes time t 2 . The value of ${{{t_2}} \over {{t_1}}} is \sqrt {{3 \over x}} $. The value of x will be _______________.

Integer / Numeric Answer Type