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

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 )

v=2*r*w



1. to find r we will divide the diameter80 cm by 2 to get the radius
80cm/2= 40cm

2. to find omega (w) we will use the formula w=wi+alpha*t

we need to use this formula to find alpha: alpha=a/r
alpha is 1/40.

so just plug into the formula w=wi+alpha*t

=0+(1/40)(2.86)
w =0.0715

so to solve the whole problem is the formula v=2*r*w

v=2*40*0.0715
=5.72 m/s

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