Angular speed of the cylinder:
A cylinder of mass 70 kg rotates from rest in frictionless bearing under the action of a mass of 7.0 kg carried by a rope wrapped around the cylinder. If the diameter is 1000 mm, what will be the angular speed of the cylinder, 2 seconds after motion starts.
Solution
Free body diagrams of the cylinder and the mass are shown in Figures (b) and (c). We should keep in mind that the tension in the rope is similar for both the diagrams.
Let the cylinder rotate in a clockwise direction such that the mass m shall move downwards. Let a be the downward acceleration of mass m and α be the angular acceleration of the cylinder. Then we may write a = rα , where r is the radius of the cylinder.
Inertia force F_{i} = m a shall be acting upwards on the mass m as shown in Figure (c).
Likewise, inertia couple M_{ i} = I_{α} shall be acting on the cylinder in the anticlockwise direction. This is shown in Figure (b).
Now we write down dynamic equilibrium equation for the mass and we get
T + ma = mg
or, T = mg - ma = 7 × 9.81 - 7a
Likewise equilibrium of moments for the cylinder will give us,
T . r - I α = 0
∴ (7 × 9.81 - 7 a) × 0.5 =( ½) × 70 × 0.5^{2} × α
∴ (68.67 - 7 × 0.5 α) 0.5 = 35 × 0.5^{2} × α
or, 34.34 - 1.75 α = 8.75 α
∴ α = 3.27 rad / sec ^{2} .
To attain angular velocity of the cylinder after 2 seconds, we simply write kinematic equation, knowing that ω_{0} = 0 .
∴ ω = ω_{0} + α t = 0 + 3.27 × 2 = 6.54 rad / sec .