1. 1.7/27 * 6.9 = 0.434 RAD/s^2.
2. V = 59 REV/min * 2pi RAD/REV * 1/60 min/s = 6.18 RAD/s.
V = a * t,
6.18 = 0.434t,
t = 6.18/0.434 = 14.2 s.
A small rubber wheel is used to drive a large pottery wheel, and they are mounted so that their circular edges touch. The small wheel has a radius of 1.7 cm and accelerates at the rate of 6.9 rad/s^2, and it is in contact with the pottery wheel (radius 27.0 cm) without slipping.
(1) Calculate the angular acceleration of the pottery wheel.
Express your answer using two significant figures.
(2) Calculate the time it takes the pottery wheel to reach its required speed of 59 rpm.
2 answers
The angular acceleration is
(1.7/27)(6.9rad/s^2)=0.434 rad/s^2
The it takes the bigger wheel to reach 59rpm
(59 rev/mn)(2Ď€rad/rev)(mn/60s)= 6.18 rad/s
w = a/t so t =w/a
t= 6.18 rad/s / 0.434rad/s^2
t = 14.2 s
(1.7/27)(6.9rad/s^2)=0.434 rad/s^2
The it takes the bigger wheel to reach 59rpm
(59 rev/mn)(2Ď€rad/rev)(mn/60s)= 6.18 rad/s
w = a/t so t =w/a
t= 6.18 rad/s / 0.434rad/s^2
t = 14.2 s
1 answer