# Need help with centripetal acceleration questions

1. Apr 19, 2004

### wikidrox

1. A cyclist accelerates from rest at a rate of 0.80 m/s^2. How fast will a point on the top of the tire rim (diameter = 68cm) be moving after 4.0 s? (hint: At any given moment, the lowest point on the tire is in contact with the ground and hence is at rest.)

2. What is the max speed with which a 1000kg car can round a turn of radius 100m on a flat road if the coefficient of friction between the tires and road is 0.5? Is the result independent of the mass of the car?

I need some assistance on how to go about doing the questions. What equations do I use where?

2. Apr 19, 2004

### Staff: Mentor

For the first problem, find out how fast the cyclist is going after 4.0 s. (Use the equations for uniform accelerated motion.) Then figure out how fast the tire must be spinning since it rolls without slipping.

For the second problem, apply Newton's 2nd law. Since the car is going around a curve, the acceleration is centripetal. The only force is friction--the maximum force of friction is μmg.

3. Apr 19, 2004

### ShawnD

First make a formula for the linear velocity of the cyclist, Vc is the velocity of the centre of the tire.

$$V_c = at$$

Now find the rotation speed of the tire, assuming rotation around where the tire touches the ground.

$$V_c = \omega r$$

$$\omega = \frac{V_c}{r}$$

Fill in the equation above (for Vc)

$$\omega = \frac{at}{r}$$

Now since you are looking at a point across the diameter of the tire, it's 2r distance away. let Va be the speed of the point at the top of the tire

$$V_a = \omega (2r)$$

$$V_a = (\frac{at}{r})(2r)$$

$$V_a = 2at$$

I'll have to think about this for a while, but I get the feeling it is not mass dependant.