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svp [43]
2 years ago
7

A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s

. Consider the merry-go-round to be a uniform disk of radius 5.5m and mass 29,000 kg. Suppose that it is supported by bearings that produce negligible friction torque.What torque specifications must the new motor satisfy?
Physics
1 answer:
pashok25 [27]2 years ago
5 0

Answer:

109656.25 Nm

Explanation:

\omega_f = Final angular velocity = 1.5 rad/s

\omega_i = Initial angular velocity = 0

\alpha = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm

The torque specifications must be 109656.25 Nm

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all I did was divide 300 and 15.

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let combined velocity be V

HENCE  

final momentum = total mass × velocity

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8 0
3 years ago
A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
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Answer:

The speed of the ball is 42.5 m/s

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Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

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V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

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