1. A small rubber wheel is used to drive a large pottery wheel. The two wheels a
ID: 1346765 • Letter: 1
Question
1.
A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The small wheel has a radius of 3.0 cm and accelerates at the rate of 6.0 rad/s2 , and it is in contact with the pottery wheel (radius 30.0 cm ) without slipping.
PART A: Calculate the angular acceleration of the pottery wheel.
PART B: Calculate the time it takes the pottery wheel to reach its required speed of 70 rpm .
2.
A person exerts a horizontal force of 58 N on the end of a door 73 cm wide.
PART A: What is the magnitude of the torque if the force is exerted perpendicular to the door?
PART B: What is the magnitude of the torque if the force is exerted at a 60.0 angle to the face of the door?
3.
A dad pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 14 rpm in 13.0 s . Assume the merry-go-round is a uniform disk of radius 2.5 m and has a mass of 560 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge.
PART A : Calculate the torque required to produce the acceleration, neglecting frictional torque.
PART B: What force is required at the edge?
Explanation / Answer
part A) linear accealration is same for the two wheels
r1*alpha1 = r2*alpha2
0.03*6 = 0.3*alpha2
alpha2 = (0.03*6)/0.3 = 0.6 rad/s^2
Part B)
wf = 70 rpm = 70*2*3.142/60 = 7.333 rad/s
wi= 0 rad/s
alpha = 0.6 rad/s^2
apply wf-wi = alpha*t
t = (wf-wi)/alpha = 7.333/0.6 = 12.22 S
2) A) T = r*F = 0.73*58 = 42.34 N-m
B) T = r*F*sin(60) = 42.34*sin(60) = 36.66 N-m
3) Tprque T = I*alpha
alpha = [(14*2*3.142/60) - 0 ]/ 13 = 0.112 rad/s^2
I = 0.5*M*R^2 +(2*m*r^2) =(0.5*560*2.5*2.5) +(2*25*2.5*2.5) = 2062.5 kg-m^2
T = 2062.5*0.112 = 231 N-m
B) T = r*F
Force F = T/r = 231/2.5 = 92.4 N
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.