Velocity & Acceleration of Bug on Rolling Wheel

[Demonsthenes]

Homework Statement

A bug of mass m is stuck to a point on the rim of a rolling wheel (of radius r)... which traces out a path called a cycloid. The position vector of the point (bug) is given by:

r(theta) = r(theta - sin(theta))i + r(1 - cos(theta))j​

For a wheel rolling with constant angular spped...

Homework Equations

A) Determine the velocity vector v = vxi + vyj.

B) Determine the acceleration vector a =axi + ayj

C) Verify the magnitude of the acceleration vector is given by a = r(greek w)^2

The Attempt at a Solution

The derivatives i took for each vector looked very incorrect.

 

[AvroArrow]

I think it's because i don't really understand what theta is. Is it the angle at which the wheel has rotated?

 

[m2003]

I would approach this problem by first understanding the physical principles involved. In this case, we have a bug stuck to a point on a rolling wheel, which means the bug is moving in a circular motion due to the rotation of the wheel. This motion can be described using the equations for velocity and acceleration in circular motion.

For the velocity vector, we can use the formula v = rw, where r is the radius of the wheel and w is the angular velocity. Since the bug is attached to the rim of the wheel, its position vector can be described by r(theta), where theta is the angle of rotation. Therefore, the velocity vector would be v = r(theta)w. Substituting the given position vector, we get v = r(theta)w = r(theta - sin(theta))wi + r(1 - cos(theta))wj.

For the acceleration vector, we can use the formula a = r(greek w)^2, where r is the radius of the wheel and w is the angular velocity. Again, substituting the given position vector, we get a = r(theta)w^2 = r(theta - sin(theta))w^2i + r(1 - cos(theta))w^2j. This is the acceleration vector in terms of the position vector and angular velocity.

To verify the magnitude of the acceleration vector, we can calculate its magnitude using the formula a = sqrt(ax^2 + ay^2), where ax and ay are the x and y components of the acceleration vector. Substituting the values from the acceleration vector we calculated, we get a = sqrt((r(theta - sin(theta))w^2)^2 + (r(1 - cos(theta))w^2)^2). Simplifying this, we get a = r(greek w)^2, which matches the given equation.

In conclusion, the velocity and acceleration vectors of the bug on the rolling wheel can be described using the equations for circular motion, and the magnitude of the acceleration vector can be verified using the given equation.