Question

the angular position of a particle revolving about an axis is given by (Theta =Q). Q(t) = t^2 - 3t +4 radian Find the acceleration of the point at time t = 2 s given radius of circular path 1m

Answer 1

Answer:

d) √5

Explanation:

In this question we have to calculate net acceleration

which is given by

A(net) = √Ac ^2 + At ^2

centripetal acceleration Ac = rw^2

= 1× 2×2 -3 = 1

tangential acceleration At = r× alpha

1 × 2 =2

A(net) = √ 1^2 + 2^2 = √5

Answer 2

The acceleration of the point at t=2 s=$\sqrt5 m/s^2$

Explanation:

Angular position$\theta=t^2-3t+4$

Radius of circular path=1 m

Angular velocity=$\omega=\frac{d\theta}{dt}$

Angular velocity=$\omega=\frac{d(t^2-3t+4)}{dt}=2t-3$

Substitute t=2

Then , $\omega=2(2)-3=1 rad/s$

Angular acceleration=$\alpha=\frac{d\omega}{dt}=\frac{d(2t-3)}{dt}$

Angular acceleration at t=2 =$2rad/s^2$

Tangential component of acceleration=$a_t=\alpha r$

Tangential component of acceleration=$2\times 1=2 m/s^2$

Normal component of acceleration=$a_n=\omega^2 r$

Normal component of acceleration=$1\times 1=1 m/s^2$

Magnitude of acceleration=$\sqrt{a^2_t+a^2_n}$

Magnitude of acceleration=$\sqrt{1^1+2^2}=\sqrt5$

Hence, the acceleration of the point at t=2 s=$\sqrt5 m/s^2$

#Learns more:

If angular displacement of a particle moving in a circular path of radius R is given as theta is equal to t²-t³.find the time when particle turns back and also the angular displacement

The answer is t is equals to 2/3 seconds and angular displacement is 4/27 radian

https://brainly.in/question/12413165: answered by Itsmahi

[The angular displacement of a particle is given by

Theta=W(Omega)t+ 1/2 alpha t^2]

Omega=1Rad/s

Alpha =1.5 Rad\s^2

angular velocity at time t=2sec will be (in Radian per second) ?????​

https://brainly.in/question/12313883: answered by brainly user