SECTION A Torque and Statics SECTIONB Rotational ...physicsatthebay.com/AP/MC_Questions_files/AP 1 H_W 3 Ans.pdf · WARNING: These areAP PhysicsCFree Response Practice Rotation ANSWERS
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4. Find the torques of each using proper signs and add up. 2 + (1) – (2) + (3) + (4) +F(3R) – (2F)(3R) + F(2R) +F(3R) = 2FR 3 1 4 B 5. Simply apply rotational equilibrium (m 1 g) • r 1 = (m 2 g) • r 2 m 1 a = m 2 b B 1. I tot = I = I 0 + I M = I 0 + M(½L) 2 A 2. = Iwhere = (3M 0 )(l) – (M 0 )(2l) = M 0 l and I = (3M 0 )(l) 2 + (M 0 )(2l) 2 = 7M 0 l 2 A 3. X = Fl; O = F O L O sin , solve for the correct combination of F O and L O C 9. = T 2 R–T 1 R = IC 10. If the cylinder is “suspended in mid air” (i.e. the linear acceleration is zero) then F = 0 D 11. = TR = I= ½ MR 2 which gives = 2T/MR and since F = 0 then T = Mg so = 2g/R the acceleration of the person’s hand is equal to the linear acceleration of the string around the rim of the cylinder a = R = 2g B 1. K tot = K rot + K trans = ½ I2 + ½ Mv 2 = ½ (2/5)MR 2 2 + ½ Mv 2 = (1/5)Mv 2 + ½ Mv 2 = (7/10)Mv 2 = MgH, solving gives H = 7v 2 /10g D 2. Mgh = K tot = K rot + K trans , however without friction, there is no torque to cause the sphere to rotate so K rot = 0 and Mgh = ½ Mv 2 A 3. Mgh = K tot = K rot + K trans = ½ I2 + ½ Mv 2 ; substituting v/r for gives Mgh = ½(I/r 2 + M)v 2 and solving for v gives v 2 = 2Mgh/(I/r 2 + M), multiplying by r 2 /r 2 gives desired answer D
