1. A giant rolls a bowling ball with a uniform speed of 30 m/s (about 70 mph) ac

Question

1. A giant rolls a bowling ball with a uniform speed of 30 m/s (about 70 mph) across the top of his large desk. The ball rolls off the edge of the desk and lands on the floor 120 m from the edge of the desk. How high is the desk? When the ball has moved a horizontal distance of 60 m how high above the floor is it?

2. A baseball is hit in a line drive and leaves the bat with a horizontal speed of 22m/s. What are its horizontal and vertical speeds 1/4 s later?

3. You are driving your race car around a circular test track. Which would have a greater effect on the magnitude of your acceleration, doubling your speed or moving to a track with half the radius? Why?

4. An Indy car with a speed of 120 km/h goes around a level, circular track with a radius of 1 km. What is the centripetal acceleration of the car (in m/s2)?

5. The maximum speed with which a 945-kg car makes a 180-degree turn is 10.0m/s. The radius of the circle through which the car is turning is 25.0 m. Determine the force of friction acting upon the car.

6. A stone tied to a string is being twirled around in uniform circular motion. If the string snaps, what will happen to the stone? Choose from the drawing below its path (represented by one of the arrows) immediately after the string breaks and explain why.

Explanation / Answer

6. When the string is left, the stone flies away tangentially. ( In a direction perpendicular to the line joining the radius to the center of the circle).

So, option C) is the correct choice.

5. We have the formula, v^2/rg= (f + tan theta)/ ( 1-f tan theta)

where f= coefecient of friction of the road.

Since tan 180= 0,

force due to friction= f mg= v^2m/r = 10 x 10 x 2 x 945/ 25 = 7560 N

4. v= 120 x 5/18 m/s =100/3 m/s

r= 1000 m

Centripetal acceleration= v^2/ r = 10^4 / (9 x 10^6)

= 1/ 900 = 0.001 m/s^2

3. Since the value of tangential acceleration remains same, net acceleration will wholly depend upon centripetal acceleration.

which is v^2/ r

doubling speed= 4v^2/ r

halving radius= 2v^2/ r

So, doubling the speed has a greater impact.

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