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rosijanka [135]
2 years ago
9

A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr

apped around the drum of radius 1.33 m exerts a force of 4.35 N to the right on the cylinder. A rope wrapped around the core of radius 0.51 m exerts a force of 6.62 N downward on the cylinder. What is the magnitude of the net torque acting on the cylinder about the rotation axis? Answer in units of N m
Physics
1 answer:
PilotLPTM [1.2K]2 years ago
3 0

Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

tau = F\times r

tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'

\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 2.41\ Nm

 

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<u>Explanation:</u>

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