a solid cylinder of 100.0 gm and a radius 4.00 cm was rolled down with a constan

ID: 1472391 • Letter: A

Question

a solid cylinder of 100.0 gm and a radius 4.00 cm was rolled down with a constant angular acceleration of 0.35 rad/s^2 with an initial speed of 10.0 rad/sec. ( rotational inertia of a solid cylinder thorugh the axis shown is (1/2)MR^2).

a) what was the angular velocity at the end of the 18 second?

b) (i) what was the total angle in radians it rotated in this 18 seconds?

(ii) how many revolutions did it complete in this time period?

c) what was the linear velocity at the end of 18th second?

d) what was the rotational kinetic energy at the end of the 18th second?

e) what was the total kinetic energy at the end of the 18th sec?

Explanation / Answer

moment of Inertia,
I = 0.5*M*R^2
= 0.5*0.1 Kg* (0.04 m)^2
= 0.002 Kg.m^2

wi = 10 rad/s
a = 0.35 rad/s^2

a)
use:
wf = wi + a* t
wf= 10 + 0.35*18
= 16.3 rad/s

b)
i)
thetha = wi*t + 0.5*a*t^2
= 10*18 + 0.5*0.35*18^2
= 236.7 rad
ii)
number of revolution = thetha/2*pi
= 236.7 / 2*3.14
= 37.7
So, it completed 37 revolution

c)
V = r*wf
=0.04 m * 16.3 rad/s
= 0.652 m/s

d)
Rotational kinetic energy = 0.5*I*wf^2
= 0.5*0.02*16.3^2
= 2.66 J

e)
translational kinetic enrgy = 0.5*m*v^2
= 0.5*0.1Kg*(0.652 m/s)^2
=0.02 J

total kinetic energy = 2.66 J + 0.02 J = 2.68 J