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A moto-cross motorcycle/rider is airborne after a jump. Each wheel of his bike has a rotational inertia of 0.760 kg m2, so the pair have twice that. He is moving at a speed of 15.0 m/s, so the tires are spinning at 50.0 radians/s. He and the motorcycle have a rotational inertia of 35.0 kg m2. In the air he panics, and clamps on the brake so the tires stop compared to the bike. What will his (bike and all) angular speed be after the wheels stop

User StephaneAG
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1 Answer

4 votes

Answer:2.17 rad/s

Step-by-step explanation:

Given

Each wheel has moment of inertia
I=0.76 kg-m^2

Linear speed speed
v=15 m/s

angular speed of tire
\omega =50 rad/s

moment of inertia of motorcycle
I_2=35 kg-m^2

Initial moment of inertia
I_1=2* 0.76=1.52 kg-m^2

Due to absence of external Torque we can Conserve Angular Moment of inertia


l_1\omega _1=I_2\cdot \omega _2


2* 0.76* 50=35* \omega _2


\omega _2=2.17\ rad/s

User Ekatz
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