First slide

Applications of SHM

Question

A solid cylinder of radius r and mass M rests on a curved path of radius R as shown figure. When displaced through a small angle $θ$ as shown and left to itself, it executes simple harmonic motion. Then the time period of oscillation is (Assume that the cylinder rolls without slipping)

Difficult

A

$T = 2 π \sqrt{(\frac{3 (R - r)}{2 g})}$
B

$T = 2 π \sqrt{(\frac{g}{(R - r)})}$
C

$T = 2 π \sqrt{(\frac{8 (R - r)}{g})}$
D

$T = 2 π \sqrt{(\frac{(R - r)}{g})}$

Solution

Restoring torque acting on a cylinder after a small displacement $θ$ , about an axis passing through point of contact O with the curved path, $τ = - M g r \sin θ$ , for small angles    $\sin θ \approx θ$
Therefore, $τ = - M g r θ$ ………….(1)
Angular acceleration of the cylinder is $α = \frac{d^{2} ϕ}{d t^{2}}$
From the diagram, $r ϕ = (R - r) θ$
$ϕ = \frac{(R - r) θ}{r}$
Consequently, $α = \frac{d^{2}}{d t^{2}} (\frac{(R - r) θ}{r})$
$α = (\frac{(R - r)}{r}) \frac{d^{2} θ}{d t^{2}}$ ……………………. (2)
Moment of inertia of the cylinder about a point of contact is $I = \frac{3}{2} M r^{2}$
Using the definition of torque $τ = I α$ and substituting equation (2) in it,
   $τ = (\frac{3}{2} M r^{2}) (\frac{(R - r)}{r}) \frac{d^{2} θ}{d t^{2}}$
Substituting equation (1) in the above relation,
$- M g r θ = (\frac{3}{2} M r^{2}) (\frac{(R - r)}{r}) \frac{d^{2} θ}{d t^{2}}$
$- M g θ = (\frac{3}{2} M) (R - r) \frac{d^{2} θ}{d t^{2}}$
$- g θ = (\frac{3}{2}) (R - r) \frac{d^{2} θ}{d t^{2}}$
This equation can be written as $0 = \frac{d^{2} θ}{d t^{2}} + (\frac{2}{3 (R - r)}) g θ$
Where $\frac{d^{2} θ}{d t^{2}} = α$ , Therefore $- (\frac{2}{3 (R - r)}) g θ = α$
Comparing this with the equation of motion for angular simple harmonic motion, $α = - ω^{2} θ$ ,
   $ω^{2} = (\frac{2 g}{3 (R - r)})$
Hence angular frequency is $ω = \sqrt{(\frac{2 g}{3 (R - r)})}$
$\frac{2 π}{T} = \sqrt{(\frac{2 g}{3 (R - r)})}$ , where    $ω = \frac{2 π}{T}$

$T = 2 π \sqrt{(\frac{3 (R - r)}{2 g})}$

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