Question

Use this fact together with Equations (1) and (2):

(3)    a = Rα =

TR2

I

=

mg − T

m

Solve for the tension T:

(4)    3.52

Substitute Equation (4) into Equation (2) and solve for a:

(5)    2

Use a = Rα and Equation (5) to solve for α:

3

MASTER IT HINTS: GETTING STARTED | I'M STUCK!

Suppose the wheel is rotated at a constant rate so that the mass has an upward speed of 4.52 m/s when it reaches a point P. At that moment, the wheel is released to rotate on its own. It starts slowing down and eventually reverses its direction due to the downward tension of the cord. What is the maximum height, h, the mass will rise above the point P?
h =

Answer 1

This is a case of uniform circular motion because the problem says that has a constant rate so that the mass has an upward speed of 4.52 m/s when it reaches a point P.

Let's assume--because the problem says that the mass is upward - that the mass is at the top of the path so the mass continues its movement in a circle.

So we have to analyse the forces acting on the mass in that point.

Hence there are two forces acting vertically on the mass one is the weight of the mass (w) and other the tension force of the wheel (Ft)

Both of the forces are acting down on a vertical axis, and the vector sum of these forces produce the centripetal acceleration to the mass in the upward point.

We will apply Newton's second law to the vertical direction.

We will choose as positive the direction down because precisely the centripetal acceleration is directed down, i.e. towards the center. Then we have:

Second law of Newton for radial movement

Fr = m ar     (Equation 1)   The sum of radial forces is equal to mass and centripetal acceleration (or also called radial acceleration)  

Therefore:

Fr = Ft + w   (Equation 2) The magnitude of the radial forces is equal to the weight of the mass (w) and the tension force of the wheel (Ft)

Matching equation 1 and equation 2:

m ar = Ft + w   where w= mg and ar (from uniform circular motion´s equation is equal to: ar = v^2 / r

Replacing in the above equation:

m v^2 / r = Ft + mg   Now given the conditions of the problem performing a physical analysis at the upward point we can match the strength of tension of the wheel to zero

So the above equation would be:

m v^2 / r = 0 + mg Cancelling out mass:

v^2 / r = g    Where the variable r is the height above the center of the wheel and is the requested outcome:

r = v^2 / g   Substituting values:

r = (4.52 m/s)^2 / 9.81 m/s^2 = 20.4304 m^2/s^2 / 9.81 m/s^2 = 2.0826 meters