Question

A clown of mass Mc= 100kg stands still at the far right end of a plank of uniform density, mass Mp= 60kg, and length L= 4m that is supported by a frictionless hinge L/4 from the right side of the plank, and it is in contact with a vertical pole hanging from the ceiling at a point that is L/4 from the left side of the plank. a) Draw an "extended” free body diagram indicating where each of the forces acts on the plank. b) What is the direction and magnitude of the force from the hinge on the plank? c) What is the magnitude of the force from the pole pushing down on the plank? d) The clown jumps from the plank into a tank of water below. Immediately after he jumps and the plank loses contact with the pole hanging from the ceiling, what is the magnitude of the angular acceleration of the plank as it begins to rotate about the frictionless hinge? Recall that the moment of inertia of a long thin uniform density object about its center of mass is Icm= (1/12)ML? e) What is the total kinetic energy of the plank right before it rotates into the vertical position? polex e clown Mc plank y me W/4 hinge L

Answer 1

Ne M9 L L mg Zelo. any point is the plank is in equilibriums. In vertical direction, the downward forces acting on the plant must be balanced by upward forres. Wpt mg+Mg = Nn NA T = + ko x 7.+1ee x ex9.8) Nu = Np + 1568 It is also in rolectional equilibrium. Hence net turque about Torque = Forue x distance of force from point. Consided Counter clockwise Torque as positive clockwise torque as negative about the Contact with pole is Zelo. Tupt Tong + Tuh + Treg o mg/2) + Nh() - Mg (84) = 0 distance of No flom pale is clockwise Eve) counter clock wire. (tre) Nh ( 2 ) = Mg(22) + mg (4) Torque co Zela Trg Ting ene is 3M9 + mg 2 = 150x9.8+ 30x9.8 2 for = 1764 1764 = Np + 1568 from 0 & ① 3 IN = 196

Moment of Inentin about Col is mg, 1 Jean-ML? 12 Inolta - m² + 16 da 129 48 e) If plank From parallel axis theorems, Momente of Inentia about hinge = Moment 나 about Cart mdz while d distance of come from lunge. TH =Inet m (4) mu (A+) - 7 ml 48 48 Tong = - IH a mg h 7 m 2²2 7 L becomes ventied then centre of mass L. The decrease in displaces vertreally by potential energy is equal to increase in rotational kinetic energy. mg 42 = £ In we mg 2 2 = 48 (24g 249 7L 7L total vinete energy is (kE= mgh سارے عام 7

The hinge applies a force of 1764 N in upwards.

The pole applies a force of 196 N in downwards.

The angular acceleration of the plank when clown jumps from the plank is 12 g / 7 L.

The total kinetic energy is rotational kinetic energy is mgL / 4.