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The angular position of a point on a rotating wheel is given by=2.0+4.0t2+2.0t3, whereis in radians andtis in seconds. At t=0s, what are (a) the point鈥檚 angular position and (b) its angular velocity? (c) What is its angular velocity at time=4.0s? (d) Calculate its angular acceleration at.t=2.0s. (e) Is its angular acceleration constant?

Short Answer

Expert verified

a. The angular position of the point att=0sis2.0rad.

b. The angular velocity of the point at t=0sis0rad/s.

c. The point鈥檚 angular velocity at t=4.0sis128rad/s.

d. The point鈥檚 angular acceleration at localid="1654588713261" t=2.0sis32rad/s2

e. The angular acceleration is not constant.

Step by step solution

01

Given data

The angular position of a point on a rotating wheel,=2.0+4.0t2+2.0t3

02

Understanding the angular velocity and angular acceleration

Angular velocity is the rate of change of displacement with time. Angular velocity is a vector quantity, with direction given by right hand rule. Angular acceleration is the time rate of change of angular velocity.

The expression for angular velocity is given as:

=诲胃dt 鈥 (i)

Here, is the angular displacement and tis the time interval.

The expression for the angular acceleration is given as:

=ddt 鈥 (ii)

03

(a) Determination of the angular position of the point att=0 s

The angular position of a point on a rotating wheel is written as,

Thus, the angular position att=0is2.0rad

04

(b) Determination of the angular velocity of the point att=0 

Using equation (i), the angular velocity of the point is calculated as:

(t)=诲胃dt=d(2.0+4.0t2+2.0t3dt=8.0t+6.0t2

Now, at t=0sthe angular velocity is,

role="math" localid="1654663354225" (t=0s)=8.0(0s)+6.0(0s)2=0rad/s

Thus, the angular velocity at is t=0is role="math" localid="1654663441148" 0rad/s

05

(c) Determination of the angular velocity of the point att=4 s

Theangularvelocityofthepointis,(t)=8.0t+6.0t2Att=4.0s,theangularvelocityis,(t=4.0s)=8.0(4.0s)+6.0(4.0s)2=32rad/s+96rad/s=128rad/s

Thus, the angular velocity of the point att=0sis128rad/s.

06

(d) Determination of the angular acceleration of the point att=2 s

Using equation (ii), the angular acceleration is calculated as:

(t)=d蝇(t)dt=d(8.0t+6.0t2)dt=8.0+12.0t

At t=2.0s, the angular acceleration is,

(t=2.0s)=8.0+12.0(2.0s)=32rad/s2

Thus, the angular acceleration is 32rad/s2.

07

(e) Determination of angular acceleration of the point is constant or not.

The equation of angular acceleration is a function of time. Hence, it is changing with respect to time. Therefore, the angular acceleration is not constant.

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