Lesson_Text_4.3

# Lesson_Text_4.3 - 1 Lesson 4.3 Problem Solving in...

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1 Lesson 4.3 Problem Solving in Rotational Motion Linear motion Rotational motion v = v o + at ϖ = ϖ o + α t x = x o +v o t + (1/2)a t 2 θ = θ + ϖ o t + (1/2) α t 2 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 (L is angular momentum) Example 1: Find the cross product of vectors A and B given that: A = 2 i – 3 j and B = -5 i + j Solution: ( 29 2 -3 2 1 ( 5 3) -5 13 1 B k k A × = - ×- = - = × - Example 2: A fixed 0.15 kg solid-disc pulley with a radius of 0.05 m is acted up on by a net torque of 6.4 m N. What is the angular acceleration of the pulley? Solution: We have: m = 0.15 kg, r = 0.05 m and τ = 6.4 m.N The moment of inertia of the disc I = (1/2)m r 2 = 0.5 x 0.15 x 0.05 2 = 1.875 x10 -4 kg.m 2 τ = I α , where α is the angular acceleration. Therefore, 4 4 2 3.4 10 . 6.4 1.875 10 r d I a s τ α - - × = = = × Example 3: A light meter stick is loaded with masses of 2.0 kg and 4.0 kg at the 30-cm and 75-cm positions respectively. (a) What is the moment of inertia about an axis through the 0-cm end of the meter stick? (b) What is the moment of

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2 inertia about an axis through the center of mass of the system? (c) Use the parallel axis theorem to find the moment of inertia about the axis through the 0-cm end of the stick, and compare the result with the result of part (a) Solution: m 1 = 2 kg m 2 = 4 kg 0 30 75 r 1 r 2 Fig. 1 (cm) x (a) Moment of inertia of the 2.0 kg about the 0-cm mark = m 1 r 1 2 = 2.0 kg x (0.3) 2 = 0.18 kg m 2 . Moment of inertia of the 4.0 kg about the 0-cm mark = m 2 r 2 2 = 4.0 kg x (0.75) 2 = 2.25 kg m 2 . Total moment of inertia of the system about the 0-cm mark = m 1 r 1 2 + m 2 r 2 2 = 0.18 + 2.25 = 2.43 kg.m 2 . (b) If the cm of the system is at the x-cm mark, taking the 0-cm mark as the origin, we can write: 1 1 2 2 1 2 2 0.3 4 0.75 0.6 6 cm m r m r x m m × + × = = = + This means that the center of mass of this system is at the 60 cm mark. The moment of inertia of the 2-kg mass about the center of mass = 2 x (0.6 – 0.3) 2 = 0.18 kg m 2 . The moment of inertia of the 4.0 kg mass about the center of mass = 4 x (0.75 – 0.6) 2 = 0.09 kg m 2 Total moment of inertia of the system about the center of mass I cm = 0.18 + 0.09 = 0.27 kg.m 2 . (c)
3 According to the parallel axis theorem, the moment of inertia about a parallel axis is given by: I = I cm + Md 2 . Where M is the total mass and d is the distance of the axis from the center of mass. Here d = 0.6 m and M = 6 kg. Therefore, I = 0.27 + 6 x 0.6 2 = 2.43 kg.m 2 .

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