Dynamics Question with Rotational Axis Theorem

[ryank614]

Homework Statement

A cyclist rides around a circular track (R= 30m) such that the point of contact of the wheel on the track moves at a constant speed of 10m/s. The bicycle is banked at 15 degrees inward from the vertical. Find the acceleration of a tack in the tire (.4m radius) as it passed through the highest point of its path. Use cylindrical unit vectors in expressing the answer

Homework Equations

acceleration = $$\ddot{P}$$ + $$\alpha$$ x P + 2 $$\omega$$ x $$\dot{P}$$ + $$\omega$$ x ($$\omega$$ x P)

Where the acceleration of the tack = acceleration of tack with respect to the second frame (the point of contact of the wheel) + angular acceleration cross product with vector from center of circular track to tack + 2 * angular velocity cross product with velocity of the tack with respect to the point of contact of wheel + $$\omega$$ x ($$\omega$$ x P)

The Attempt at a Solution

I know this problem is a matter of finding all the unknowns. I have found $$\omega$$ and angular acceleration.

I think I can also find OP (we will call O the center of the track, so this is the vector to the tack).

How do you find the velocity and acceleration of the tack with respect to the point of contact of the wheel on track?

 

[BigJDubs]

Once I have all the knowns, I can plug it into the acceleration equation. \omega = \frac{V}{R} = \frac{10}{30} = 0.333 rad/sAlpha = \frac{\omega^{2}}{R} = (0.333)^2/30 = 0.00555 rad/s^2OP = \sqrt{R^2 - (0.4)^2} = 29.6m Velocity of tack with respect to point of contact of wheel on track = ? Acceleration of tack with respect to point of contact of wheel on track = ? Then, acceleration of tack = \ddot{P} + \alpha x P + 2 \omega x \dot{P} + \omega x (\omega x P) = ?