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Angelina_Jolie [31]
4 years ago
9

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A

constant tangential force of 250 N applied to the edge causes the wheel to have an angular acceleration of 0.940 rad/s^2.
(a) What is the moment of inertia of the wheel?
(b) What is the mass of the wheel?
(c) If the wheel starts from rest, what is its angular velocity after 5.00 s have elapsed, assuming the force is acting during that time?
Physics
1 answer:
Ira Lisetskai [31]4 years ago
3 0

Answer:

(a) I = 87.766 kg.m²

(b) m = 1611.864 kg

(c) ω = 4.7 rad/s

Explanation:

Given;

radius of the wheel, r = 0.330 m

applied force, F = 250 N

angular acceleration, α = 0.940 rad/s²

<u>Part (a)</u> the moment of inertia of the wheel

τ = F x r = Iα

I = (F x r) / α

where;

I is the moment of inertia

I = (250 x 0.33) / 0.94

I = 87.766 kg.m²

<u>Part (b)</u> the mass of the wheel

I = ¹/₂mr²

2I = mr²

m = (2I) / r²

where;

m is the mass of the wheel

m = (2 x 87.766) / (0.33²)

m = 1611.864 kg

<u>Part (c) </u>angular velocity after 5.00 s

ω = ω₀ + αt

where;

ω is the angular velocity after 5.00 s

ω₀ is the initial angular velocity = 0

ω = 0 + (0.94 x 5)

ω = 0 + 4.7

ω = 4.7 rad/s

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Read 2 more answers
You are rushing to the train station to catch your morning commute. The train leaves the train station from rest with an acceler
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Answer:

a) t= 4.81 s,and t= 23.51.

b)d=6.88 m

c)v=4.8 m/s

Explanation:

Acceleration of train ,a= 0.6 m/s²

u = 0 m/s

Your speed ,V= 8.5 m/s

Lets take after t time he you catch the train

Distance travel by train in t time

d=ut+\dfrac{1}{2}at^2

d=\dfrac{1}{2}\times 0.6\times t^2  ----------1

d= V ( t- 4)

d= 8.5 ( t- 4)  --------2

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0.3 t² =  8.5 ( t- 4)

0.3 t² -8.5 t + 34 = 0

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It means that you will catch train after t= 4.81 s,and t= 23.51.

t=23.51 sec means that you will catch the train after 23.51 sec also because acceleration of train is low.

Distance travel

d= 8.5 ( t- 4)

t= 4.81 s

d= 8.5 ( 4.81- 4)  m

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Lets speed = v

0.3 t² =  v ( t- 4)

0.3 t² - v t + 4 v = 0

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D=\sqrt{b^2-4ac}

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v = 4 x 0.3 x 4

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