Need help with centripetal acceleration questions

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To solve the centripetal acceleration questions, the first problem involves calculating the speed of a cyclist after 4 seconds, using the formula Vc = at, where a is the acceleration (0.80 m/s²) and t is time. The speed of the point on the top of the tire can be determined as Va = 2at, which accounts for the tire's rotation. For the second problem, the maximum speed of a 1000kg car rounding a turn with a radius of 100m can be found using the frictional force, expressed as μmg, where μ is the coefficient of friction (0.5). The analysis suggests that the maximum speed is independent of the car's mass, focusing instead on the frictional force and radius of the turn. Understanding these principles is crucial for accurately solving centripetal acceleration problems.
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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?
 
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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.
 
wikidrox said:
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.)

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


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