Question

A 51-cm-diameter wheel accelerates uniformly about its center from 150 rpm to 290 rpm in 4.0 s.(A) Determine it's angular acceleration. (B) Determine the radial component of the linear acceleration of a point on the edge of the wheel 1.1 s after it has started accelerating. (C) Determine the tangential component of the linear acceleration of a point on the edge of a wheel 1.1 s after it has started accelerating.

Answer 1

Answer:

(A) 7/6 pi /s^2

(B) 4.145 m/s^2

(C) 119pi/200 m/s^2

Explanation:

Part A.

The angular acceleration is given by

$\alpha=\frac{\omega_f-\omega_i}{\Delta t}$

where wf is the initial angular velocity and wi is the final angular velocity and t is the time interval.

Now, we are given the angular velocity is given in rpm and we have to convert it into radians/sec .

150 rpm = 150 x 2pi / 60 min = 5 pi rad/ sec

290 rpm = 290 x 2pi / 60 min = 29/ 3 pi rad/ sec

Now we are in the position to find the angular acceleration

$\alpha=\frac{\frac{29}{3}\pi-5\pi}{4s-0s}$$\boxed{\alpha=\frac{7}{6}\pi\; /s^2}$

which is our answer!

Part B.

The radial acceleration is given by

$a_r=\frac{v^2}{R}$

where v is the velocity of the object (moving in a circle) and R is the radius of the circle.

Now,

$v=\alpha Rt$

putting in the values of alpha, R and t = 1.1 s gives

$v=\frac{7}{6}\pi\times\frac{0.51}{2}\times1.1$$v=1.028\; m/s$

therefore,

$a_r=\frac{v^2}{R}=\frac{(1.028)^2}{0.51/2}$$\boxed{a_r=4.145/s^2}$

which is our answer!

Part C.

Here we have to relationship between angular and tangential acceleration:

$a=\alpha R$

where r is the radius of the circle.

Since R = 0.51/2 m, we have

$a=\frac{7}{6}\pi\cdot\frac{0.51}{2}m$

$\boxed{a=\frac{119}{400}\pi}$

which is our answer!

Hence, to summerise

(A) 7/6 pi /s^2

(B) 4.145 m/s^2

(C) 119pi/200 m/s^2