Biomechanics / Physics problem: Torque. See below. Help please!?

Question

Q. What amount of force must be applied by the elbow flexors inserting at an average perpendicular distance of 0.015m from the axis of rotation at the elbow over a period of 0.3seconds to stop the motion of the 3.5kg lower arm swinging downwards with an angular velocity of 5.0rad/s when the radius pf gyration for the lower arm is 0.20m? Prove by calculating.

Possible solutions:
1) T = F * d and F = ma, for a use g (9.81m/s/s)
2) T = F * d, and F = ma, for a use 5rad/s / 0.3s
3) T.t = Δ I ω and I = mr²

Which one is correct?! Thanks.

Answer 1

solution no2

Answer 2

If the question extremely states "on the doorways hinge" then the answer ought to should be 0 because torque is rigidity cases lever arm, and on the door's hinge, it truly is the pivot element, the lever arm equals 0, meaning the torque = 0. in case you mean on the door cope with, the answer should be seventy seven*cos(30)*.57. it really is because the lever arm is .57 (on account that i anticipate both meters is the proper of the refrigerator) and the 30 comes from 100 and twenty-ninety because for torque, the rigidity and lever arm should be perpendicular (it truly is why I subtracted the ninety ranges)

Answer 3

I don't think any of your possible solutions are correct.

The way I would solve this is as follows:

Force = rate of change of momentum
force*perp. distance= change in angular momentum/ time

Where the
change in angular momentum = I w

There are formulas for the interia(I) of generic shapes such as 1/2MR^2, I suggest you look up the appropriate one.

and Force*distance=torque.

p.s. I hope this is correct, good luck.

Answer 4

Using the third equation:

Torque needed to stop
T = 3.5 x 0.2 x 0.2 x 5 / 0.3 = 0.23 Nm

Using the equation: T = F*d.

F x 0.015 = 0.23 Nm.

F = 15.6 N.