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Newton’s 2nd Law for Rotation
Quiz - Two wheels, start from rest with forces applied as shown. Assume that all of the mass is in the rims, so the moment of inertia I = mr2. In order to have identical angular accelerations, how large must F2 be? A = 0.5 N, B = 1 N, C = 2 N, D = 4 N Clicker channel = 44
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Topics for Today Newton’s second law for rotation (10-7)
Work and rotational kinetic energy (10-8) Rolling = Translation + Rotation (11-1)
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Wheels Torque 𝜏=𝑟𝐹 sin 𝜙 , where 𝜙 is the angle between 𝑟 and 𝐹 .
Second law is 𝜏=𝐼𝛼 Small wheel 𝜏 1 = 𝑅 1 𝐹 1 = 𝐼 1 𝛼 1 = 𝑚 1 𝑅 1 2 𝛼 1 ⇒ 𝛼 1 = 𝐹 1 𝑚 1 𝑅 1 Same equation for 𝛼 2 , setting 𝛼 2 = 𝛼 1 = 𝐹 2 𝑚 2 𝑅 2 = 𝐹 1 𝑚 1 𝑅 1 𝐹 2 = 𝐹 1 𝑚 2 𝑅 2 𝑚 1 𝑅 1 = 1 𝑁 1 𝑚 0.5 𝑚 =2 𝑁
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Work and Rotational Kinetic Energy
Work produces a change in kinetic energy Δ𝐾=𝑊= 𝐹𝑑𝑥 This holds for rotational kinetic energy Δ𝐾= 𝐼 𝜔 𝑓 2 − 𝐼 𝜔 𝑖 2 = 𝑊 To calculate rotational work, we use angular displacement and torque 𝑊= 𝜃 𝑖 𝜃 𝑓 𝜏𝑑𝜃 Since 𝑠=𝑟𝜃, we have 𝑑𝑠=𝑟𝑑𝜃, therefore 𝑑𝑊=𝐹𝑑𝑠=𝐹𝑟𝑑𝜃=𝜏𝑑𝜃 Rotational power 𝑃= 𝑑𝑊 𝑑𝑡 =𝜏 𝑑𝜃 𝑑𝑡 =𝜏𝜔
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Work and Rotational Kinetic Energy
Do demo 1Q20.10, angular momentum machine with the masses at 2R and R. Keep string with weight at fixed radius. Quiz – what is the ratio of the final kinetic energy with the masses at 2R to that with R. A=1:4, B=1:2, C=1:1, D=2:1, E=4:1 Quiz – what is the ratio of the final angular velocities with the masses at 2R to that with R.
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Work and Rotational Kinetic Energy
Do demo 1Q20.10, angular momentum machine with the masses fixed. Put string at R and 2R. Quiz – what is the ratio of the torque with the string at 2R to that with R. A=1:4, B=1:2, C=1:1, D=2:1, E=4:1 Quiz – what is the ratio of the final kinetic energy with the string at 2R to that with R.
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Rolling = Translation + Rotation
Assuming the center of mass of the bicycle and rider are moving at +5 m/s (relative to the ground), how fast is the bottom edge of the wheel moving (relative to the ground)? +5 m/s 0 m/s -5 m/s +10 m/s
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