A disc of mass m is connected to an ideal spring of force constant 'k'. If disc is released from rest, then what is maximum friction force on disc (in N)? Assuming friction is sufficient for pure rolling

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TestHub · Accepted Answer

Released from rest at the natural length; maximum extension from work-energy: $m g x_{m} sin θ = \frac{1}{2} k x_{m}^{2} \Rightarrow x_{m} = \frac{2 m g s i n θ}{k}$

At the lowest point, let friction be f up the incline: $k x_{m} - m g sin θ - f = M a$

Pure rolling: $f R = \frac{1}{2} M R^{2} \frac{a}{R} \Rightarrow f = \frac{M a}{2}$

$2 m g sin θ - m g sin θ = M a + \frac{M a}{2} = \frac{3 M a}{2} \Rightarrow a = \frac{2 g s i n θ}{3}$

$f = \frac{M a}{2} = \frac{M g s i n θ}{3} = \frac{1 \times 10 \times 0.6}{3} = 2 N$

Answer: 2

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