A 4.70 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The?

A 4.70 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The spool is a uniform solid cylinder of radius 6.00 cm and mass 2.40 kg.





Figure 10.20

(a) What is the net torque on the system about the point O?

Magnitude



Direction



(b) When the counterweight has a speed v, the pulley has an angular speed = v/R. Determine the total angular momentum of the system about O.



(c) Using the fact that = dL/dt and your result from (b), calculate the acceleration of the counterweight.

A 4.70 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The?
a) Torque = Force * Distance

Torque = mg*R

Torque = (4.70 * 9.81 * .06)

For direction, look at the diagram (and remember the right hand rule)



b) L = I*w

w = v/R

I = (1/2)(M)(R^2) + mR^2

(here we are treating the block as a particle orbiting around the axis O)

L = (M/2 + m)*R*V

L = (2.4/2 + 4.7)*.06*V



c) Torque = dL/dt

since L = (I*V/R)

T = (I*A/R)

we have torque from part a)

and we have I/R from part b)

so (T*R/I) = A

A = (answer from part a / answer from part b)

gert