1) A cyclist accelerates from rest at a rate of 1.00 m/s2. How fast will a point

Question

1) A cyclist accelerates from rest at a rate of 1.00 m/s2. How fast will a point on the rim of the tire (diameter = 80.0 cm) at the top be moving after 2.86 s? (image http://imgur.com/SPSgsun ) 2) Calculate the moment of inertia of a 12.7 kg sphere of radius 0.621 m when the axis of rotation is through its center. 3) A teenager pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 14.2 rpm in 11.8 s. Assume the merry-go-round is a uniform disk of radius 2.44 m and has a mass of 740 kg, and two children (each with a mass of 23.8 kg) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque. What force is required at the edge? 4) image ( http://imgur.com/j0F2NHp) The wheel shown above has a moment of inertia of 0.27 kg

Explanation / Answer

The net torque on the wheel = (28*0.24 ) - (18*0.24) - ( 35*0.12) = -1.8 N-m

moment of inertia = 0.27 kg m^2

so angular acceleraion = torque / inertia = -1.8 / 0.27 = -6.66667 rad /sec2

so..

angular accelation a = -6.666667 rad / sec2

intial velocity (u) = 0

final angular velocity (v) = -850 rad / sec

so time = (v-u) / a = ( -850 / (-6.6667)) = 127.5 secs

part B

kinetic energy = 0.5 * inertia * ( v^2 - u^2 )

= 0.5 * 0.27 * ( 850^2 - 0^2 ) = 97,537.5 Joules

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