Question

A long, uniform rod of length  0.530 mm and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.35 rad/srad/s and a moment of inertia about the axis of 3.30×10⁻³ kg⋅m2kg⋅m2 . An insect initially standing on the rod at the axis of rotation decides to walk to the other end of the rod. When the insect has reached the end of the rod and sits there, its tangential speed is 0.160 m/sm/s . The insect can be treated as a point mass.

A) Find the mass of the rod?

B) Find the mass of the insect?

Answer 1

2 0.35 rad/s w I = 3.3X163 kam 0.53 mm لم of when the insect is at axis rotation dislance the imect axis roation foom и 3e90 This Moment i zelo Moment Inertia rod about ily of a end by inerhia ow is bry given I = 1/3 me? 2 3.3x10 3 = 1/2 M (0.53) ML - O,0 35 24 М. kg 35.24g (An A) cs

B Insect reaches the end ng wat =0:16 Ve=WL 0-16 0.53 T = 0.30 vacy sunce here 15 no external Torque will be conserred (L ) Angular momemlüm So Li = 나 2 Liz = 1 / 3 M22 w 0:03524 *(0.53) W 3 3.3 x 10²x0.35 - mass og m + ML ų 3) w insect- (0.5373) ( 13 m2? 4 = (3.34103 R 0.30 Now, Li= 3.3810 X163*0.35 = (3.3816 (3.38162 m m (0-S2)) 0.2 196 16² kg ma 1.96810 21-96g (dans & cs