[nj_bartel]

I have two problems regarding rotational motion.

1) Use conservation of energy to determine the angular speed of the spool shown in Figure P8.36 after the 3.00 kg bucket has fallen 4.35 m, starting from rest. The light string attached to the bucket is wrapped around the spool and does not slip as it unwinds.


I initially set it up to solve it as a conservation of energy problem by setting PE = mgh = (3)(9.81)(4.35) = 128.0205

I then equated this to (1/2)(m_bucket)(v_{final(bucket)} + (1/2)(I_spool(w_spool²)

I then was unsure where to proceed, so if anyone could help me out that'd be great.

I did however see another means of solving it (or so I think) not using conservation of energy.

I solved for the torque of the spool by (0.6)(3)(9.81) = (r)(F_bucket) = 17.658

I then solved for the angular acceleration by 17.658 = (5)(0.6)²(a) and got a = 9.81

Since the circumference = 2(pi)(r), I got the circumference = 2(pi)(0.6) = that many meters / rotation

The bucket fell 4.35m, so that = 1.153873337 rotations, which = 7.25 radians, the change in theta.

I then used the equation w² = w_i²(a)(change in theta) to get (2)(9.81)(7.25) = w²

So w = 11.92665, however this answer was wrong (by less than 10%).







2) A potter's wheel having a radius 0.48 m and a moment of inertia 12.7 kg·m2 is rotating freely at 47 rev/min. The potter can stop the wheel in 5.5 s by pressing a wet rag against the rim and exerting a radially inward force of 70 N. Find the effective coefficient of kinetic friction between the wheel and the wet rag.

I solved for a by dividing w_i by the time, then multiplied a by I to get the force of friction, -5.455175723.

I then set u_k = (F_fric)(F_N) = (-5.455175723) / (-70) = .07793, but this was wrong.




Any help would be greatly appreciated, thanks!

Nick

[Borek]

Conservation of energy should work, just find the relation between bucket linear speed and spool rotational speed.

[nj_bartel]

v_bucket = the tangential velocity of the spool?

so v_bucket = (0.6)(w)


Then (1/2)(m_bucket)(0.6w)² + (1/2)(I_spool)(w_spool)² = 128.0205

(1/2)(3.00)(0.6w)² + (1/2)(0.9)(w_spool)² = 128.0205

1.98w² = 256.041

w = 11.3716



That came out right, thanks 


Still need help with the other one if anyone can

[Borek]

I solved for a by dividing w_i by the time, then multiplied a by I to get the force of friction, -5.455175723.

Shouldn't you use a moment of force (torque)?

[nj_bartel]

torque = mr²a

Is that the correct formula, or should it be torque = Ia?