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 260 n applied to its edge causes the wheel to have an angular acceleration of 0.810 rad/s2. (a) what is the moment of inertia of the wheel?
1 answer:
Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
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Answer:
Induced current, I = 18.88 A
Explanation:
It is given that,
Number of turns, N = 78
Radius of the circular coil, r = 34 cm = 0.34 m
Magnetic field changes from 2.4 T to 0.4 T in 2 s.
Resistance of the coil, R = 1.5 ohms
We need to find the magnitude of the induced current in the coil. The induced emf is given by :
Where
is the rate of change of magnetic flux,
And
Using Ohm's law,
Induced current,
I = 18.88 A
So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.