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### Course: Physics library>Unit 4

Lesson 2: Centripetal forces

# Bowling ball in vertical loop

In this video David explains how to find the normal force on a bowling ball rolling in a vertical loop. Created by David SantoPietro.

## Want to join the conversation?

• At , David says that the speed of the ball in the outermost right position will increase compared to the speed in the top position. And in the case of vertical circular motion, it seems quite intuitive to me. But when we derived formulas for centripetal acceleration in earlier videos, we assumed that the speed is constant. Can you explain why we are still allowed to use these formulas in this case? Thank you!
• When we derive the formulas for centripetal acceleration, we are only looking at a single force at a time. The centripetal force will not increase the speed of the ball by itself; it will only change the direction of the velocity just enough to make the ball travel in a circular path. If there are no other forces than the centripetal force, the speed will be constant. Therefore, we don't have to assume that the speed of the object is constant, only that the speed of the object is unaffected by the centripetal force.
In this example, however, we also have to consider the force of gravity. At any given moment, the centripetal force is equal to Fn + Fg ⋅ sin(x). When x = 180 degrees, the force of gravity does not contribute to the centripetal force. This does not mean that gravity doesn’t accelerate the ball, it only means that gravity doesn’t change the direction of the velocity. If it did, the ball couldn’t travel in a circular path anymore.
• What if the bowling ball was somewhere at an angle? Would gravity affect it?
(putting a clock for comparison) if the bowling ball was at 10 o'clock (which is between 12 o' clock and 9 o'clock) it would be at an angle...
• Gravity will always be accelerating the ball by 9.8 m/s² downward no matter where the ball is. What changes is the normal force of the track pushing on the ball as it moves through the loop. Normal force will be greatest at the bottom of the loop, smallest at the top, and somewhere in between those two values based on the angle of the centrifugal force + gravitational force to the surface.
• Isn't the ball also exerting a force on the loop?
• Yes, that is the reaction force to the normal force, so it is equal to the normal force. The force diagram does not include the force that the ball is exerting on the loop because the force diagram only shows what forces are exerting on the ball.
• I'm a bit confused; since at the top of the circle mg is acting downwards, why is there a normal force there in the first place? Should not (acc. to Newton's 3rd Law) the force exerted on the bowling track (Which is what we are looking for) be the pairing force with mg?
• It depends how fast the ball is going. If the ball is going very fast, mg will not be sufficient force to keep the ball on the circular path, so the track will have to do some pushing as well. The way we find the minimum required speed is to ask how slow can the ball go before the normal force becomes zero, which is when the ball is just losing contact with the track (i.e. "falling")
• How do you calculate the speed of the ball when it is at the outermost right or left side of the loop?
• In the video, you already know what the normal force is so you can equal it to mv¨2/r. Or, you could calculate v with energies, ET at the top= PE+KE, at the bottom ET= KE, and depending on how position changes, PE will change. Let's say PE decreases, KE will have to increase because energy is conserved around the loop (assuming it's a perfect circle)
• In case of rotating a string , is the normal force replaced by tension? Are their magnitude also equal?
• yes, the centripetal force is provided by tension instead of a normal force.
In other instances it might be provided by gravity, or by an electromagnetic force.
• Why do we need to include the normal force when calculating the centripetal force? Wouldn't the normal force get cancel out by the force exerted by the ball on the loop since it is an action-reaction pair?
• I hope I'm not misleading you now. It seems to me that 3rd law and action-reaction pairs are taken into account when dealing with a System of two or more bodies. Acc. to 2nd law, change in motion is due to External net force. Yes, the ball may communicate force upward at the top, i.e. on the loop de loop, but this force is not external, does not applies to the ball's motion. What loop experiences is of no interest; hopefully it's rigid.
(1 vote)
• wait, isn't the normal force always perpendicular to the force of gravity? If all the forces are pointing toward the center of the circle (gravity, normal force) then why doesn't the ball just do that: go towards the center of the circle, maybe even fall towards the center?
(1 vote)
• No, the normal force is perpendicular to the surface at the point of contact. For example it you have a ramp the normal force is only in the direction perpendicular to the ramp's surface, this only opposes a portion of the force of gravity and that is why things will tend to slid down a ramp if not stopped by friction.