How Is φ(t) Related to θ(t) in the Rolling Wheel Inside a Cylinder Problem?

[ranger1716]

wondering if someone could shed some light on this problem.



a wheel of radius r rolls around the interior of a cylinder of radius R. assume that the center of the cylinder is at the origin and at time t=0, the point of tangency is at the point (R,0). let P denote the original point of tangency on the wheel. we will investigate the motion of this point P on hte wheel. let vector v sub 1 (t) denote the vector emanating from the center of the wheel and ket theta(t) denote the angle v sub 1 (t) makes with the x-axis at time t. let vector v sub 2 (t) be the vector that emanates at the center of the wheel and terminates at P and let phi(t) be the angle that vector v sub2(t) makes with the horizontal at time t.


show that phi(t)=((R-r)/r)(theta(t))

I'm having trouble beginning this proof.

Any help would be appreciated.

 

[Aaron Bex]

To prove this, you will need to consider the relationship between v1(t) and v2(t). Notice that v1(t) is a vector with its origin at the center of the wheel, and its end point at the point of tangency of the wheel and the cylinder. Similarly, v2(t) is a vector with its origin at the center of the wheel and its end point at the point P.

Since the wheel is rolling around the interior of the cylinder, the point of tangency is always moving in a circular motion. This means that the angle theta(t) that v1(t) makes with the x-axis at any given time t is changing. Additionally, since the point P is always located on the circumference of the wheel, it is also following a circular motion, with the angle phi(t) that v2(t) makes with the x-axis at any given time t also changing.

Now, consider a right triangle with sides of length r, R-r, and R. Since the wheel is rolling around the interior of the cylinder, the angle θ(t) that v1(t) makes with the x-axis must be equal to the angle that the side of length R-r makes with the side of length R. This implies that the angle θ(t) that v1(t) makes with