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

[underoath0101]

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

 

[ehild]

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

 

[tomcue]

I would approach this problem by first defining the variables and equations involved. The speed of the gum is determined by its velocity, which is the rate of change of its position over time. The velocity can be calculated using the equation v = d/t, where v is velocity, d is distance, and t is time.

(a) To determine the number of revolutions the wheel makes in a second, we can use the equation for circumference, C = 2πr, where r is the radius of the wheel. In this case, r = 0.5 meters. We know that the center of the wheel advances by 31.4 meters in one second, so we can set up the equation 31.4 = 2π(0.5)n, where n is the number of revolutions. Solving for n, we get n = 31.4/2π(0.5) ≈ 10 revolutions.

(b) To determine the force on the gum, we can use Newton's second law, F = ma, where F is force, m is mass, and a is acceleration. In this case, the gum has a mass of 20.0 grams, which is equivalent to 0.02 kilograms. The force acting on the gum is the centripetal force, which is given by F = mv^2/r, where v is the velocity and r is the radius of the wheel. From part (a), we know that the wheel makes 10 revolutions in one second, so the velocity of the gum is v = (10)(2πr)/t, where t is the time for one revolution. Since the train is moving at a constant speed, the time for one revolution is the same as the time for one second, so t = 1 second. Plugging in the values, we get v = (10)(2π)(0.5)/1 = 10π ≈ 31.4 meters/second. Now we can calculate the force on the gum using F = (0.02)(31.4)^2/(0.5) ≈ 390.5 Newtons.

(c) The speed of the gum as seen by the center of the wheel is the same as the speed of the wheel, since the gum is stuck to the rim of the wheel. Therefore, the speed of the gum as seen by the center of the wheel is also approximately 31