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:
See the explanation below
Explanation:
We must solve this problem by defining that when we have a constant velocity, the acceleration is equal to zero. That is, when there is no speed change, there is no acceleration. We can understand it very easily by means of the following equation of kinematics.
where:
Vf = final velocity = 55 [km/h]
Vo = initial velocity = 55 [km/h]
a = acceleration [m/s²]
t = time [s]
As we can see there is no change in speed, and the difference between the two is equal to zero.