Question

A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How many radians did the disk turn while stopping ? C. how many revolutions? ​

Answer 1

Answer:  

A. α = - 1.047 rad/s²  

B. θ = 14.1 rad  

C. θ = 2.24 rev  

Explanation:  

A.  

We can use the first equation of motion to find the acceleration:

$\omega_f = \omega_i + \alpha t$  

where,  

ωf = final angular speed = 0 rad/s  

ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s  

t = time = 3 s  

α = angular acceleration = ?  

Therefore,

$0\ rad/s = 3.14\ rad/s + \alpha(3\ s)$  

α = - 1.047 rad/s²

B.  

We can use the second equation of motion to find the angular distance:

$\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2$  

θ = 14.1 rad

C.  

θ = (14.1 rad)(1 rev/2π rad)  

θ = 2.24 rev