Lesson_4.3_Printable

# Lesson_4.3_Printable - Problem solving in Rotational Motion...

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Problem solving in Rotational Motion In order to solve problems in rotational motion you need to recall all the formulae we developed earlier. It is easier to write down the formulae for linear motion and then write the corresponding equation for rotational motion. Linear motion Rotational motion ϖ = ϖ o + α t v = v o + at = x v + (1/2)a t 2 + (1/2) 2 x o +v o t + (1/2)a t θ = θ o + ϖ o t + (1/2) α t v 2 = v o 2 + 2 a x ϖ 2 = ϖ o 2 + 2 α θ F = m a τ = I α W = F x W rot = τ ⋅ θ K = (1/2)m v 2 K rot = (1/2) I ϖ 2 P = m v L = I ϖ 2 2 1 1 2 2 rolling cm K I mv ϖ = + 1

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A 3.5 kg cylinder of radius 15 cm is initially at rest. A rope of negligible mass is wrapped around it and pulled with a force of 17 N. Find (a) the torque exerted by the rope, (b) the angular acceleration of the cylinder and (c) the angular velocity of the cylinder after 5.0 s. m = 3.5 kg, r = 15 cm = 0.15 m, F = 17 N (a) F = 17 N is at right angles to r = 0.15 m τ = r . F sin θ = r. F = 0.15 × 17 = 2.55 mN 0.15 m (b) To find α , we need the moment of inertia I of the cylinder. A cylinder is the same as a disc of radius r Moment of inertia of a cylinder is I = (1/2)m r 2 = 0.5 × 3.5 × 0.15 2 = 0.04 kg.m 2 τ = I α F = 17 N I τ α = 2 2.55 m.N 0.04 kg.m = -2 63.75 rad.s = 2
(c) the angular velocity of the cylinder after 5.0 s. (c) ϖ o = 0, t = 5 s ϖ f = ϖ o + α t = 0 + 63.75 × 5 = 318.75 rad s -1 0.15 m F = 17 N 3

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A wheel mounted on an axle that is not frictionless is initially at rest. A constant external torque of 50 N is applied to the wheel for 20 s giving it an angular velocity of 600 rpm. The external torque is then removed and the wheel comes to rest in 120 s. Find (a) the moment of inertia of the wheel, (b) the frictional torque that brings the wheel to a stop. (a) ϖ o = 0, ϖ f = 600 rpm = (600 × 2 π )/60 = 20 π rad.s -1 We first find the angular acceleration α τ = 50 m.N, t = 20 s Now we use the relation τ = I α and solve for I I τ α = f o t ϖ - = ϖ f = ϖ o + α t -2 50 m.N rad.s π = -1 20 rad.s 20 s = -2 rad.s = 2 50 kg.m = 4
ϖ o = 600 rpm = 20 π rad.s -1 , ϖ f = 0, t = 120 s and I = 50/ π kg.m 2 . (b) When the external torque is removed the frictional torque τ f produces a negative angular acceleration bringing the wheel to a stop We find the acceleration caused by τ f using the relation: f τ = I α f o t ϖ α - = ϖ f = ϖ o + α t -1 0 20 rad.s 120 s π - = -2 rad.s 6 = - 2 -2 50 = kg.m rad.s 6 3 m N = - 5

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## This note was uploaded on 05/01/2011 for the course PHY 2048 taught by Professor George during the Fall '10 term at Edison State College.

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Lesson_4.3_Printable - Problem solving in Rotational Motion...

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