A disk with mass m = 8 kg and radius R = 0.33 m begins at rest and accelerates u

Question

A disk with mass m = 8 kg and radius R = 0.33 m begins at rest and accelerates uniformly for t = 17.8 s, to a final angular speed of omega = 34 rad/s. What is the angular acceleration of the disk? Rad/s^2 What is the angular displacement over the 17.8 s? rad What is the moment of inertia of the disk? Kg-m^2 What is the change in rotational energy of the disk? J What is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s^2 What is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s^2 What is the final speed of a point on the disk half-way between the center of the disk and the rim? m/s What is the total distance a point on the rim of the disk travels during the 17.8 seconds? m

Explanation / Answer

Ans :-

1. = Wf –Wi /t = 34/17.8 =1.91rad/s^2

2. = Wi t + ½ *t^2

       = 0 + ½ * 1.91 *(17.8)^2

= 302.58rad

3 I= ½*mr^2 = ½*8*(0.33)^2 = 0.4356kgm^2

4.change in rotational kinetic energy = ½ IWf^2 - ½ IWi^2 = ½ *0.4356 *34^2 = 251.78J

5.Wf = 17rad /s

= 17/8.9 = 1.91

a= *r = 1.91*0.33 = 0.6303m/s^2

6 ar= 17^2 * 0.33 = 95.37m/s^2

7]v= w*r = 17*0.33 =5.61m/s

8 ] x = v0*t + ½ *at^2

X= 0 +1/2 0.6303*17.8^2

X = 99.85m

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