Reshma said:
Linear speed v = \omega R
So if T is the time period, distance in one revolution = circumference
vT = 2\pi R
Please note that the motion here is downhill(clockwise) so there will be changes in the sign if we take into account the direction. So how do I make these changes?
remember that linear velocity (
v), linear acceleration (
a), angular velocity (
ω), and angular acceleration (
α) are all
vector quantities. when solving a problem, one of the first jobs is to select convenient (orthogonal) coordinate axes into which these vector quantities can be projected into their coordinate components.
for your problem, it's typical to consider the ramp with highest level on the left, lowest level on the right, with motion from "left-to-right". then it's also typical to establish a coordinate system consisting of:
1) "x" axis parallel to ramp surface, (+) direction left-to-right
2) "z" axis normal (perpendicular) to ramp surface, (+) direction upward from ramp surface
3) "y" axis into page (normal to both above axes), (+) direction into page
for your problem, the linear motion is "left-to-right" along the "x" axis and thus its x-component is (+). the angular motion is clockwise, which according to the "right-hand rule", produces a (+) "y" component.
thus, for this problem, all quantities have (+) values for their components.
with the above in mind, and working with the aformentioned components, you basically have solved the problem.
beginning where you left off:
If T is the period of revolution, distance in 1 revolution = 1 circumference:
vT \ = \ 2\pi R
v \ = \ (2\pi /T) R
v \ = \ (\omega) R
\frac{dv}{dt} \ = \ a \ = \ \frac{d(\omega R)}{dt} \ = \ \frac{d\omega}{dt} \cdot R \ = \ \alpha R