1) What is the acceleration of the hanging mass ( in m/s^2) while the wheel\'s s
ID: 1374593 • Letter: 1
Question
1) What is the acceleration of the hanging mass ( in m/s^2) while the wheel's speed is increasing?
2) What is the tension in the string (in N) while the wheel's speed is increasing? (Should be two forces)
3) Determine the torque exerted on the wheel by the string while the wheel's speed is incresing
4) Determine the moment of inertia of the wheel. You may assume here that the only significant torque acting on the wheel is that due to the string. (i.e., that the torue due to the frictional force is negligible.)
Consider a bicycle wheel with a radius of 30 cm and 16 spokes that is mounted with its axle fixed in a horizontal position. A 50-g mass is hung from a string wrapped around the periphery of the tire. The wheel is held stationary with the weight hanging as shown and then released. The wheel starts to spin and spins faster and faster until the string slips off 5 seconds after release. An electronic device counts the number of spokes that pass in front of it and sends the information to a computer. This allows us to obtain a measurement of the angular speed of the wheel. The readings of the display device are summarized in the graph below. 1) What is the acceleration of the hanging mass ( in m/s^2) while the wheel's speed is increasing? 2) What is the tension in the string (in N) while the wheel's speed is increasing? (Should be two forces) 3) Determine the torque exerted on the wheel by the string while the wheel's speed is incresing 4) Determine the moment of inertia of the wheel. You may assume here that the only significant torque acting on the wheel is that due to the string. (i.e., that the torue due to the frictional force is negligible.)Explanation / Answer
initial nagular speed, w1 = 0
final angular speed, w2 = (10/16)*2*pi rad/s
= 3.925 rad/s
a) angular acceration of the wheel = (w2-w1)/t
= (3.925-0)/5
= 0.785 rad/s^2
accelaration of the ahanging mass, a_tan = r*alfa
= 0.3*785
= 0.236 m/s^2
b) Initail tension, T = m*g = 0.05*9.8 = 0.49 N
final tension, T = m*g - m*a
= 0.05*9.8 - 0.05*0.236
= 0.4782 N
c) Torque = T*r*sin(90)
= 0.4782*0.3*1
= 0.143 N.m
d) Torque = I*alfa
==> I = Torque/alfa
= 0.143/0.785
= 0.183 kg.m^2
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