A bicycle rim has a diameter of 0.65 m and a moment of inertia, measured about its center, of 0.21 kg⋅m2What is the mass of the
1 answer:
Answer:
m = 1.99 kg = 2 kg
Explanation:
The moment of inertia of a bicycle rim about it's center is given by the following formula:
where,
I = Moment of Inertia of the Bicycle Rim = 0.21 kg.m²
r = Radius of the Bicycle Rim = Diameter of the Bicycle Rim/2
r = 0.65 m/2 = 0.325 m
m = Mass of the Bicycle Rim = ?
Therefore,
<u>m = 1.99 kg = 2 kg</u>
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Refer to the diagram shown below.
Let ω = the angular velocity (rad/s) of the wheel.
At the topmost point, the passenger, with mass m, will feel weightless if the passenger's weight matches the centripetal force.
The passenger's weight is mg.
The centripetal force is mω²r.
Therefore
mω²r = mg
ω = √(g/r)
Because g = 9.8 m/s² and r = 25/2 = 12.5 m, therefore
ω = √(9.8/12.5) = 0.8854 rad/s
Because 1 revolution per second is 2π rad/s, therefore
Answer: 8.455 rev/min
Let ω = the angular velocity (rad/s) of the wheel.
At the topmost point, the passenger, with mass m, will feel weightless if the passenger's weight matches the centripetal force.
The passenger's weight is mg.
The centripetal force is mω²r.
Therefore
mω²r = mg
ω = √(g/r)
Because g = 9.8 m/s² and r = 25/2 = 12.5 m, therefore
ω = √(9.8/12.5) = 0.8854 rad/s
Because 1 revolution per second is 2π rad/s, therefore
Answer: 8.455 rev/min