the angular position of a point on a rotating wheel is given by 741 168t2 222t3 where is in radi

(a) The point's angular position at t = 0 is given by substituting t = 0 into the equation for θ

The angular position of a point on a rotating wheel is given...

The angular position of a point on a rotating wheel is given by Î¸ = 7.41 + 1.68t^2 + 2.22t^3, where Î¸ is in radians and t is in seconds. At t = 0, what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at t = 5.02 s? (d) Calculate its angular acceleration at t = 2.63 s. (e) Is its angular acceleration constant?

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00:01 Hey everyone, welcome back.

00:02 In this particular question we are given with angular position of a particle which is changing with respect to time and this is given as theta is equals to 7 .41 plus 1 .68 into t squared plus 2 .22 into tq.

00:19 This is the angular position of a particle and this is changing with respect to time.

00:24 Now in the first part of question we have to find out its angular position at t equals to 0.

00:29 Theta at t is equals to 0 so for this i will simply have to put the value of t as 0 then theta not which is the angular position at time t is equals to 0 this is coming out as 7 .41 radiance 7 .41 radian now in the second part of question we are asked to find out what exactly is the angular velocity so angular velocity is denoted by change in position with respect to time change in angular position with respect to time is defined as angular velocity.

01:03 So if i differentiate this particular function then this is coming out as 2 times of 1 .68 into t plus 3 times of 2 .22 into t squared.

01:14 This is differentiation of this particular function.

01:17 So in this in the same part also we have we are asked to find out what exactly is the value of angular velocity at t equals to 0.

01:26 So if i put t equals to 0 in this particular equation then this is coming out as 0.

01:30 0 radian per second.

01:32 0 radian per second.

01:34 This is the angular velocity as t equals to 0.

01:37 Now in the same part we are asked to find out, in the next part we are asked to find out what exactly is the angular velocity at at time t is equals to 5 .02 seconds.

01:50 So in this question, in this part we have to put t is equals to 5 .02 in this particular equation.

01:56 Okay, so omega will come out as 2 into 1 .6 8.

02:00 Into time which is 5 .02 plus 3 into 2 .22 into 5 .02 whole square okay so omega will come out as 184 .7018 radian per second this is the angular velocity at time t is equals to 5 .0 2 second now in the next part of question we are asked to find out what exactly is the angular acceleration so angular acceleration is nothing but rate of change of angular velocity with respect to time...

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