60 cm diameter disk has a mass of 2 kg rotating in a horizontal plane (constant

Question

60 cm diameter disk has a mass of 2 kg rotating in a horizontal plane (constant angular acceleration = 1.5 rad/s^2. Imagine a bug of 20 grams is sitting near the rim of a disk. When the disk starts to rotate, the bug has an angular position of 59 degrees Find the total moment of inertia of the disk and the bug, You can find this by finding two separate moments of inertia and add together. When t=3.5 seconds: i. Find the angular velocity i. Find Linear velocity and linear acceleration of bug. ili. Find rotational kinetic energy of the disk and of the bug (both have A. B. iv. v. vi. the same angular velocity Angular position of the bug Centripetal accel of bug Frictional force wheel is exerting on bug

Explanation / Answer

A)

moment of inertia I = Idisk + Ibug

I = (1/2)*Mdisk*Rdisk^2 + Mbug*Rdisk^2

I = (1/2)*2*0.3^2 + 0.02*0.3^2

I = 0.0918 kg m^2

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B)

angular velocity w = wo + alph*t

w = 0 + 1.5*3.5 = 5.25 rad/s

(ii)

Linear velocity v = R*w = 0.3*5.25 = 1.575 m/s

linear acceleration a = R*alpha = 0.3*1.5 = 0.45 m/s^2

(iii)

rotational kinetic energy K = (1/)*I*w^2 = (1/2)*0.0918*5.25^2 = 1.265 J

(iv)

angular position

theta_0 = 59 deg = 59*(2pi/360) = 1.03 rad

theta - theta_o = (w + wo)*t/2

theta - 1.03 = 5.25*3.5/2

theta = 10.22 rad = 10.22/2pi = 1.63 revolution

angular position = 0.63*360 = 227 degrees

v)

centipetal acceleration a = R*w^2 = 0.3*5.25^2 = 8.27 m/s^2

vi)

frictional force F = Mbug*a = 0.02*8.27 = 0.1654 N

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