This one seems like it should be easy, not sure where the trouble is.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

A circular ice rink lies in a horizontal plane. A puck of mass M is propelled from point A along the rail of the ice rink so that the puck moves in a circular path. The magnitude of the initial tangential velocity is [tex]v_0[/tex]. The rail exerts a frictional force [tex]\mu F_c[/tex] on the puck causing the velocity [tex]v(t)[/tex] to decrease with time, [tex]t[/tex]. The magnitude of the centripeal force is [tex]F_c[/tex] and [tex]\mu[/tex] is the coefficient of friction between the puck and the rail. The radius of the ice rink is [tex]R[/tex]. Assume there is no friction between the puck and the ice.

a) Calculate the speed [tex]v(t)[/tex] of the puck.

b) Calculate the total distance the puck will travel from t=0 to [tex]t=\infty[/tex], i.e. [tex]s=\int_0^\infty v(t)dt[/tex]

2. Relevant equations

[tex]a_c=\frac{v(t)^2}{R}[/tex]

3. The attempt at a solution

[tex]F_f=\mu F_c=\mu M a_c = \frac{\mu M v^2}{R}[/tex]

[tex]\frac{dv}{dt}=-a_f=-\frac{F_f}{M}=-\frac{\mu v^2}{R}[/tex]

[tex]\frac{dv}{v^2}=-\frac{\mu dt}{R}[/tex]

[tex]\int_{v0}^{v}\frac{dv}{v^2}=-\int_0^t\frac{\mu dt}{R}[/tex]

[tex]\frac{1}v-\frac{1}{v_0}=\frac{\mu t}{R}[/tex]

[tex]v=\frac{R v_0}{R+\mu v_0 t}[/tex]

Now this result must be incorrect, because

[tex]s=\int_0^{\infty}{\frac{R v_0}{R+\mu v_0 t}dt}=\frac{R}{\mu}\ln(R+\mu v_0 t)=\infty[/tex] ... nonsense

Where did I go wrong?

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# Homework Help: [PhD Qualifier] Hockey puck friction

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