Question 448: Centre of Mass and Rotational Motion

Question

A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be $${1 \over 2}\sqrt {xgh} $$ m/s. The value of x is ___________.

Accepted Answer

For rolling wheel
$$[12g\sin \alpha  - 3g\sin \alpha ] 	imes R = (2 	imes 12{R^2} + 3{R^2}) 	imes {a \over R}$$
$$ \Rightarrow {{9g\sin \alpha } \over {27}} = a$$
$$ \Rightarrow a = {{g\sin \alpha } \over 3}$$
$$	herefore$$ $$v = \sqrt {2 	imes {{g\sin \alpha } \over 3} 	imes {h \over {\sin \alpha }}}  = \sqrt {{2 \over 3}gh} $$
$$ = {1 \over 2} 	imes \sqrt {{8 \over 3}gh} $$
$$	herefore$$ $$x = {8 \over 3} = 2.67$$

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type