What is the speed of the gum as seen by the center of the wheel?

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SUMMARY

The discussion focuses on the dynamics of a train wheel and the forces acting on a piece of chewing gum attached to it. A train moves at a speed of 31.4 meters per second, and the wheel has a radius of 0.5 meters. The wheel makes 62.8 revolutions per second, calculated from the distance traveled. The force on the gum is determined using the formula F=mv²/r, where m is the mass of the gum (20.0 grams) and v is the tangential speed of the gum as it moves in a circular path.

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underoath0101
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4. A train is moving with a speed of 31.4 meters
second . One of the major wheels of the locomotive has a radius of 0.5 meters.
(a) How many revolutions does the wheel make in a second if the center of the wheel advances by 31.4 meters?
(b) A piece of chewing gum, weighing 20.0 grams, is stuck near the rim of the wheel. Determine the force on the gum.
(c) What is the speed of the gum as seen by the center of the wheel?
(d) What is the velocity of the gum (when it is at the top of the wheel) as seen by a person standing on the station?

I was able to solve a, but I keep getting confused on B..b/c i know once you have B you can do F=mv^2/r to get V..idk please help
 
Physics news on Phys.org
The chewing gum moves together with the wheel, and it is near the rim, so it travels along a circle with radius R. What is the resultant force needed for uniform circular motion?

ehild
 

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