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

Topics for Today Newton’s second law for rotation (10-7) Work and rotational kinetic energy (10-8) Rolling = Translation + Rotation (11-1)

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

Work and Rotational Kinetic Energy Work produces a change in kinetic energy Δ𝐾=𝑊= 𝐹𝑑𝑥 This holds for rotational kinetic energy Δ𝐾= 1 2 𝐼 𝜔 𝑓 2 − 1 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

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.

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.

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