How to Calculate Tension and Acceleration for a String Wrapped Around a Hoop?

[MAPgirl23]

A string is wrapped several times around the rim of a small hoop with
radius r = 7.90x10−2 m and mass 0.175 kg. The free end of the string
is pulled upward in just the right way so that the hoop does not move
vertically as the string unwinds.

a) Find the tension in the string as the string unwinds
** here I used the equation for I = MR^2

I = (0.175 kg)(7.90×10−2 m)^2 = 1.092×10−3 now how do I find
the tension?

b) Find the angular acceleration of the hoop as the string unwinds.

c) Find the upward acceleration of the hand that pulls on the free end of
the string.

Please help.

 

[OlderDan]

A string is wrapped several times around the rim of a small hoop with
radius r = 7.90x10−2 m and mass 0.175 kg. The free end of the string
is pulled upward in just the right way so that the hoop does not move
vertically as the string unwinds.

a) Find the tension in the string as the string unwinds
** here I used the equation for I = MR^2

I = (0.175 kg)(7.90×10−2 m)^2 = 1.092×10−3 now how do I find
the tension?

b) Find the angular acceleration of the hoop as the string unwinds.

c) Find the upward acceleration of the hand that pulls on the free end of
the string.

Please help.

From my reply to this same problem elsewhere. You obviously recognize some of this but so I don't have to retype it

Newton's second law requires that the acceleration of an object be proportional to the net applied force. Rigid bodies, like this hoop, are complicated because different parts of the object are moving and accelerating in different directions. Fortunately, a detailed analysis of the problem reveals that you can treat a rigid body in terms of the linear motion and the rotational motion separately. Linear acceleration of the center of mass of the object is proportional to the net appplied force, and angular acceleration about the center of mass is proportional to the net torque about the center of mass. You need to look at these separately

F = Ma
T = I(alpha)

recognizing that a relationship exists between one of the forces in the problem and the net torque.

 

[thepatientmental]

a) To find the tension in the string, we can use the equation for rotational motion: τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. Since the hoop is not moving vertically, the torque is equal to zero. We can rearrange the equation to solve for tension as follows:

τ = Iα
τ = (0.175 kg)(7.90×10−2 m)^2 α
0 = (0.175 kg)(7.90×10−2 m)^2 α
α = 0

Since the angular acceleration is equal to zero, the tension in the string is also equal to zero. This means that the string is not experiencing any force while the hoop is not moving vertically.

b) Since the hoop is not moving vertically, the only force acting on it is the tension in the string, which we have already determined to be zero. Therefore, the net torque on the hoop is also equal to zero. This means that the hoop will not experience any angular acceleration and will remain at rest.

c) The upward acceleration of the hand can be found using the equation F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, the force is equal to the tension in the string, which we have already determined to be zero. Therefore, the upward acceleration of the hand is also equal to zero. This means that the hand will not experience any upward acceleration while pulling on the string in this specific scenario.