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InterviewSolution
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A particle is moving on a circular path of radius `1.5 m` at a constant angular acceleration of `2 rad//s^(2)`. At the instant `t=0`, angular speed is `60//pi` rpm. What are its angular speed, angular displacement, linear velocity, tangential acceleration and normal acceleration and normal acceleration at the instant `t=2 s`. |
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Answer» Initial angular speed is given in rpm (revolution per minute). It is expressed in `rad//s` as `1 rp m=(2pi rad)/(60 s)` `omega_(o)=(60/pi)xx(2pi rad)/(60 s)=2 rad//s` At the instant `t=2s`, angular speed `omega_(2)` and angular displacement `theta_(2)` are calculated by using eq. `omega_(2)=omega_(0)+alphat` Substituting values `omega_(0)=2 rad//s, alpha=2 rad//s^(2), t=2 s`, we have `omega_(2)=6 rad//s` Substituting values `theta_(o)=0` rad, `omega_(o)=2 rad//s, omega=6 rad//s` `t=2 s`, we have `theta_(2)=8 rad` Linear velocity at `t=2 s`, can be calculated by using eq. `v_(2)=r omega_(2)` Substituting `r=1.5 m` and `omega_(2)=6 rad//s`, we have `v_(2)=9 m//s` Tangential acceleration `a_(tau)` and normal acceleration `a_(n)` can be calculate by using eq. and respectively. `a_(tau)=ralpha` Substituting `r=1.5 m` and `alpha=2 rad//s^(2)`, we have `a_(tau)=3 m//s^(2)` `a_(n)=omega^(2)r` substituting `omega_(2)=6 rad//s` and `r=1.5 m`, we have `a_(n)=54 m//s^(2)` |
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