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confuted
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This one seems like it should be easy, not sure where the trouble is.
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 v_0. The rail exerts a frictional force \mu F_c on the puck causing the velocity v(t) to decrease with time, t. The magnitude of the centripeal force is F_c and \mu is the coefficient of friction between the puck and the rail. The radius of the ice rink is R. Assume there is no friction between the puck and the ice.
a) Calculate the speed v(t) of the puck.
b) Calculate the total distance the puck will travel from t=0 to t=\infty, i.e. s=\int_0^\infty v(t)dt
a_c=\frac{v(t)^2}{R}
F_f=\mu F_c=\mu M a_c = \frac{\mu M v^2}{R}
\frac{dv}{dt}=-a_f=-\frac{F_f}{M}=-\frac{\mu v^2}{R}
\frac{dv}{v^2}=-\frac{\mu dt}{R}
\int_{v0}^{v}\frac{dv}{v^2}=-\int_0^t\frac{\mu dt}{R}
\frac{1}v-\frac{1}{v_0}=\frac{\mu t}{R}
v=\frac{R v_0}{R+\mu v_0 t}
Now this result must be incorrect, because
s=\int_0^{\infty}{\frac{R v_0}{R+\mu v_0 t}dt}=\frac{R}{\mu}\ln(R+\mu v_0 t)=\infty ... nonsense
Where did I go wrong?
Homework Statement
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 v_0. The rail exerts a frictional force \mu F_c on the puck causing the velocity v(t) to decrease with time, t. The magnitude of the centripeal force is F_c and \mu is the coefficient of friction between the puck and the rail. The radius of the ice rink is R. Assume there is no friction between the puck and the ice.
a) Calculate the speed v(t) of the puck.
b) Calculate the total distance the puck will travel from t=0 to t=\infty, i.e. s=\int_0^\infty v(t)dt
Homework Equations
a_c=\frac{v(t)^2}{R}
The Attempt at a Solution
F_f=\mu F_c=\mu M a_c = \frac{\mu M v^2}{R}
\frac{dv}{dt}=-a_f=-\frac{F_f}{M}=-\frac{\mu v^2}{R}
\frac{dv}{v^2}=-\frac{\mu dt}{R}
\int_{v0}^{v}\frac{dv}{v^2}=-\int_0^t\frac{\mu dt}{R}
\frac{1}v-\frac{1}{v_0}=\frac{\mu t}{R}
v=\frac{R v_0}{R+\mu v_0 t}
Now this result must be incorrect, because
s=\int_0^{\infty}{\frac{R v_0}{R+\mu v_0 t}dt}=\frac{R}{\mu}\ln(R+\mu v_0 t)=\infty ... nonsense
Where did I go wrong?
