A 1.00-kg wheel is tied to a string and allowed to roll around in a circle of ra
ID: 2240377 • Letter: A
Question
A 1.00-kg wheel is tied to a string and allowed to roll around in a circle of
radius R = 0.33 m on a horizontal table. The other end of the string passes
through a hole in the center of the table, and a counterweight of equal mass is
tied to it. The suspended object remains in equilibrium while the wheel on the
table top revolves. If the coefficient of static friction between the wheel and
the table is 0.35, what is minimum and maximum speed of the wheel?
A 1.00-kg wheel is tied to a string and allowed to roll around in a circle of radius R = 0.33 m on a horizontal table. The other end of the string passes through a hole in the center of the table, and a counterweight of equal mass is tied to it. The suspended object remains in equilibrium while the wheel on the table top revolves. If the coefficient of static friction between the wheel and the table is 0.35, what is minimum and maximum speed of the wheel?Explanation / Answer
There are two forces that affect the value of the centripetal force on the rolling wheel:
1. the weight of the hanging mass (mg = 1kg*9.8m/s^2 = 9.8 N)
2. the friction force between the wheel and the table (Ff = mu*N = 0.35*1kg*9.8m/s^2) = 3.43 N
The maximum centripetal force = 9.8N + 3.43N = 13.23N
The minimum centripetal force = 9.8N - 3.43N = 6.37N
The maximum speed is given by mv^2 / r = 13.23. Solving for v we have: V(max) = 2.1 m/s
The minimum speed: mv^2 / r = 6.37, therefore V(min) = 1.4 m/s
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