Explanation Given: Mass of yo-yo, M = 400 k g Radius of yo-yo, r = 1.5 m Axle’s radius, R=0 .1 m Formula used: Second equation of motion: v 2 = v o 2 + 2 a Δ h Where, v is the final velocity, v o is the initial velocity, a is the acceleration and Δ h is the displacement. Newton’s second law of motion: ∑ F = m a Where, m is the mass and a is the acceleration. Torque can be obtained by: τ = r × F Where, F is the applied force and r is the distance from the rotational axis. Calculation: The yo-yo’s final speed, its acceleration and fall distance is related by following equation v 2 = v o 2 + 2 a △ h Since v o = 0 Therefore, v 2 = 2 a Δ h ... ( 1 ) Applying Newton’s second law on the yo-yo ∑ F x = m g − T . .. ( 2 ) ∑ τ 0 = T r ∑ τ 0 = I 0 α ... ( 3 ) Here, I 0 = m r 2 2 is moment of inertia But a = R α ... ( 4 ) R is the radius and α is the angular acceleration

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ISBN: 9781429206099

Author: Tipler

Publisher: MAC HIGHER

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Chapter 9, Problem 85P

To determine

Linear speed of the Yo-Yo at the end of the fall.

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Chapter 9, Problem 84P

Chapter 9, Problem 86P

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PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS

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