Question

The structure shown here is made from six pieces attached as shown. Each piece has the same mass, M. The pieces are as follows: One solid, uniform disk with diameter D Four identical thin uniform rods of length D, connected to the disk as shown (opposing pairs of rods aligned with the disk center) One thin, uniform ring connected to the outer ends of the four rods, as shown. The entire structure is laid flat, at rest, on a frictionless surface and pinned to the surface with a frictionless pin at Point P. (So the diagram is an overhead view.) Point P is located at the midpoint of one of the rods, as shown. Point X is located on the edge of the disk-directly opposite to point P-as shown. Then at time t 0, a steady torque, τ, is applied to the structure around point P. and as a result, the structure begins to rotate around that pin. rod ring rod disk rod rod a. Find an algebraic expression for a), the magnitude of the acceleration of point X, as a function of time, t. b. Find an algebraic expression for W(t), the total work done on the structure, as a function of time, t Each expression may contain only known values (and the variable 1). These are known values: M, D,T.

Answer 1

(a) Moment of inertia of structure about the centre of the disc = I_G and I_G = 1/2 M (D/2)^2 + 4 times (13 MD^2/12) + M (3 D/2)^2 = MD^2 [1/8 + 13/3 + 9/4] = MD^2 [3 + 104 + 54/24] = 161/24 MD^2 so moment of inertia as structure about point P = 161 MD^2/24 + 6MD^2 = I = MD^2 [161 + 144/24] = 305MD^2/24 so the angular acceleration about point P = alpha = tau/I = 24 tau/305 MD^2 and so a_x (t) = 3/2 D alpha = 3/2 D [24 tau/305 MD^2] a_x (t) = 36 tau/305 MD (b) angle described by the structure during time t = theta = 1/2 alpha t^2 so w(t) = tau theta = tau times 1/2 times 24 tau/305 MD^2 t^2 or w(t) = 12 tau^2/305 MD^2 t^2 w (t) =

Answer 2

Let a, be the radial acceleration, then at point x it is, 3d 2 Let ar is the tangential acceleration then at point x it is given as, 3d The magnitude of acceleration at point x is, 3d 2 3d 2I

Work done is, 2I Here in I in both parts (a) and (b) is the moment of inertia at point P on axis It is given by applying parallel axis theorem 3D 1 D 12 161 . 2 MID 24 161 . 2 Substitute-MD, for ICM in equation 1=1cr +6MD2 24 161 2 24 305 MD 24