First slide

Rolling motion

Question

A solid cylinder rolls up an inclined plane of inclination $θ$ with an initial velocity v. How far does the cylinder go up the plane?

Moderate

A

$\frac{3 v^{2}}{2 g \sin θ}$
B

$\frac{v^{2}}{4 g \sin θ}$
C

$\frac{3 v^{2}}{g \sin θ}$
D

$\frac{3 v^{2}}{4 g \sin θ}$

Solution

Let the cylinder go up the plane upto a height h. Let M and R be the mass and radius of cylinder respectively. According to law of conservation of mechanical energy, we get
$\frac{1}{2} {Mv}^{2} + \frac{1}{2} {Iω}^{2} = Mgh$

( $∵$ For a solid cylinder, I = $\frac{1}{2} {MR}^{2}$ )
$\frac{1}{2} M v^{2} + \frac{1}{4} M R^{2} ω^{2} = M g h$
$\frac{1}{2} {Mv}^{2} + \frac{1}{4} {Mv}^{2} = Mgh (∵ v = Rω)$
$\frac{3}{4} {Mv}^{2} = Mgh$
$h = \frac{3 v^{2}}{4 g}$
Let s be distance travelled by the cylinder up the plane.
$s = \frac{h}{S i n θ} = \frac{3 v^{2}}{4 g S i n θ}$

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