Question

A disk (radius 95.8 cm] rotates about a fixed axis through its center of mass and perpendicular to the disk. At t = 0 the disk is rotating at frequency 6.47 Hz, and it accelerates uniformly to frequency 73.8 Hz after spinning through 38.4 revolutions. Find the angular acceleration of the disk, in rad/s².

The wheel of a cart has radius 53.9 cm. How many revolutions will it make if the cart goes 2.79 km?

An electric fan is turned off, and it slows down from 639 revolutions/s to 30 revolutions/s in 1.94 seconds. How many times did the fan blades spin in this time?

Answer 1

1.

From third law of rotational kinematics law,

wf^2 - wi^2 = 2**
here, wf = final angular speed = 2*pi*f_final

wi = initial angular speed = 2*pi*f_initial

= angular displacement = 38.4 revolutions = 38.4*2*pi rad

= angular acceleration = ??

given, f_final = final frequency = 73.8 Hz

f_initial = initial frequency = 6.47 Hz

So,

(2*pi*f_final)^2 - (2*pi*f_initial)^2 = 2**
(2*pi*73.8)^2 - (2*pi*6.47)^2 = 2**(38.4*2*pi)

= [(2*pi*73.8)^2 - (2*pi*6.47)^2]/(2*38.4*2*pi)

= 442.16 rad/sec^2

2.

We know that:

Angular displacement () = d/R

d = linear distance travelled = 2.79 km = 2790 m

R = radius of wheel = 53.9 cm = 0.539 m

So,

= 2790/0.539 = 5176.25 rad

Since 1 rev = 2*pi rad, So

= (5176.25 rad)*(1 rev/(2*pi rad))

= 5176.25/(2*pi) = 823.826 rev

= 824 revolumtions

3.

Using 1st rotational kinematic equation:

wf = wi + *t

So, = (wf - wi)/t

wi = 639 rev/s & wf = 30 rev/s, and t = 1.94 sec, So

= (30 - 639)/1.94 = -313.92 rev/s^2

Now Using 3rd rotational kinematic equation:

wf^2 = wi^2 + 2**

= (wf^2 - wi^2)/(2*)

= (30^2 - 639^2)/(2*(-313.92))

= 648.92 rev = 649 rev

Let me know if you've any query.