Question

In figure whell `A` of radius `r_A = 10 cm` is coupled by belt `B` to whell `C` of radius `r_C = 25 cm`. The angular speed of whell `A` is increased from rest at a constant rate of `1.6 rad//s^2`. Find the time needed for whell `C` to reach an angular speed of `12.8 rad//s`, assuming the belt does not slip.
.
A. 15 s
B. 12.5 s
C. 20 s
D. 10 s

Answer 1

Correct Answer - C
( c) If the belt does not slip, the linear speeds at the two rims must be equal. Since the belt does not slip, a point on the rim of wheel `C` has the same tangential acceleration as a point on the rim of wheel `A`. This means that `prop_A r_A = prop_C r_C`. where `prop_A` is the angular accleration of wheel `A` and `prop_c` is the angular acceleration of wheel `C` . Thus,
`prop_C =((r_A)/(r_C)) prop_c = ((10 cm)/(25 cm))(1.6 rad//s^2) = 0.64 rad//s^2`
With the angular speed of wheel `C` given by `omega_C = prop_C t`,
The time for it to reach an angular speed of `omega = 100 rev//min = 10.5 rad//s` starting from rest is
`t = (omega_C)/(prop_C) = (12.8 rad//s)/(0.64 rad//s^2) = 20 s`.