InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A angular positio of a point on the rim of a rotating wheel is given by `theta=4t-3t^(2)+t^(3)` where `theta` is in radiuans and `t` is in seconds. What are the angualr velocities at (a).`t=2.0` and (b). `t=4.0s` (c). What is the average angular acceleration for the time interval that begins at `t=2.0s` and ends at `t=4.0s`? (d). What are the instantaneous angular acceleration at the biginning and the end of this time interval? |
|
Answer» Angular velocity `omega=(dtheta)/(dt)=(d)/(dt)(4t-3t^(2)+t^(3))` or `omega=4-6t+3t^(2)` (a). At `t=2.0s,omega=4-6xx2+3(2)^(2)` or `omega=4rad//s` (b). At `t=4.0s,omega=4-6xx5+3(4)^(2)` or `omega=28rad//s` (c). Average angular acceleration `alpha_(av)=(omega_(f)-omega_(i))/(t_(f)-t_(i))=(28-4)/(4-2)` or `alpha_(av)=12rad//s^(2)` (d). Instantaneous angular acceleartion is , `alpha=(domega)/(dt)=(d)/(dt)(4-6t+3t^(2))` or `alpha=-6+6t` at `t=2.0s` `alpha=-6+6xx2=6rad//s^(2)` at `t=4.0s` `alpha=-6+6xx4=18rad//s` |
|