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UNO [17]
1 year ago
10

Please help me with this rotation and linear motion problem

Physics
1 answer:
kirill115 [55]1 year ago
5 0

The initial angular velocity of the wheel would be 15.5 rad/s

<h3>What is Velocity?</h3>

The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second.

The relation between the linear velocity and the angular velocity

V= ωr

as given the initial  linear velocity is 8.22 m/s

the radius of the wheel is 0.530 m

By substituting the value of the linear velocity and the radius

8.22 = 0.530ω

ω = 8.22/0.530

ω = 15.5 rad/s

Thus, the initial angular velocity of the wheel comes out to be  15.5 rad/s

Learn more about Velocity from here

brainly.com/question/18084516

#SPJ1

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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
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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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