- #1
RoyalCat
- 670
- 2
Homework Statement
A mass is released from rest atop an incline angled at Θ relative to the horizon, from a certain distance up the incline (Not height), x_0. This point up the incline is denoted P.
The mass then travels along a horizontal path of length S.
The mass then goes up an incline angled at Θ relative to the horizon.
All surfaces are smooth, the layout looks something like this:
\______/
What is the exact value of x so that T, the period of the motion (The time it takes the mass to return to the point P), is minimal?
The text-book answer is \frac{S}{2}
I keep getting \frac{S}{4}
I've also double-checked the book's answer, and the period is shorter for the book's answer.
Homework Equations
Kinematic equations.
The Attempt at a Solution
One period of motion consists of the mass going down the incline, traveling along the distance S, going up the opposite incline, going down the opposite incline, traveling back along the distance S and going back up the first incline.
The two motions along the distance S are identical, as are all 4 of the motions of the mass going up/down the inclines (As energy is conserved).
Therefore, the period is:
T=4t_1+2t_2
Where t_1 is the time it takes the mass to travel down the incline, or up the incline, and t_2 is the time it takes the mass to travel the distance S.
For the motion down the incline, with x measured along the incline from the point P:
x(t)=\frac{g\sin{\theta}}{2}t_1^{2}
x(t_1)=x_0
x_0=\frac{g\sin{\theta}}{2}t_1^{2}
t_1=\sqrt{\frac{2x_0}{g\sin{\theta}}}
The speed at the bottom of the incline is:
v_s=\sqrt{2g\sin{\theta}x_0}
x(t)=v_st
x(t_2)=S
t_2=\frac{S}{v_s}=\frac{S}{\sqrt{2g\sin{\theta}x_0}}
T=4t_1+2t_2
T(x_0)=4\sqrt{\frac{2x_0}{g\sin{\theta}}}+2\frac{S}{\sqrt{2g\sin{\theta}x_0}}
z\equiv\sqrt{x_0}
T(z)=(4\sqrt{\frac{2}{g\sin{\theta}}})z+(2\frac{S}{\sqrt{2g\sin{\theta}}})\frac{1}{z}
T'(z)=4\sqrt{\frac{2}{g\sin{\theta}}}-(2\frac{S}{\sqrt{2g\sin{\theta}}})\frac{1}{z^{2}}
T'(z_{min})=0
4\sqrt{\frac{2}{g\sin{\theta}}}=(2\frac{S}{\sqrt{2g\sin{\theta}}})\frac{1}{z^{2}}
\frac{1}{z^{2}}=(4\sqrt{\frac{2}{g\sin{\theta}}})(\frac{\sqrt{2g\sin{\theta}}}{2S})
\frac{1}{z^{2}}=4*\frac{1}{2S}*2=\frac{4}{S}
x_{0_{min}}=z_{min}^{2}=\frac{S}{4}
Bleh, could someone please read through it and tell me where I went wrong?
Thanks in advance, Anatoli.
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