A wooden rod is freely hanging from a hook in the ceiling, and the rod is covere

Question

A wooden rod is freely hanging from a hook in the ceiling, and the rod is covered in flypaper. A fly, traveling horizontally, lands on the flypaper rod and sticks there. Which statement is true? the angular momentum of the system (fly and rod) is conserved about the hook because only gravity is acting on the system. the angular momentum of the system (fly and rod) is conserved about the hook because neither the hook nor gravity exerts any torque on this system about the hook. both the angular momentum of the system (fly and rod) and its kinetic energy are conserved. both the linear momentum and the angular momentum of the system (fly and rod) are conserved. the angular momentum of the system (fly and rod) is not conserved because the fly exerts a force on the rod. A dune buggy is traveling south at 5.0 m/s. The mass of the buggy's tire, 5.0 kg, is uniformly distributed along the rim, which has a radius of 20 cm. What are the magnitude and direction of the angular momentum of the tire about its axis? 5.0 kg m^2/s vertically upwards 2.0 kgm^2/s towards the east 5.0 kgm^2/s towards the west 2.0 kgm^2/s towards the west 5.0 kgm^2/s towards the east If point P begins on the x-axis at rest, and starts rotating at the angular acceleration shown below, what is its tangential speed (not the angular speed) when it reaches the y-axis? 0.35m/s. 0.49 m/s. 0.71m/s. 0.18m/s.

Explanation / Answer

1)

   The angular momentum of the system (fly and rod) is conserved about the hook because only gravity is acting on the system.

2)

   Angular momentum of the tyre is L = mvr sin theta, here theta is 90 deress

   tyre moving toward south about its axis , L = 5*5*0.2 sin 90 = 5 kg m2/s

   the direction is in v X r direction, that is upward

option A is answer

3)
   given radius r = 2 m, alpha angular acceleration = 0.019 rad/s2, v = ?

   we know that v= r* omega

   distance covered , reaching y axis is 2 pi*r/4 = pi*r/2 = pi*2/2 = pi= 3.14 m

   from equaions of motion theta = w_0*t + 1/2 *alpha*t^2

               pi = 0+ 1/2 *0.019*t^2 ==> t = 18.184 s

           now v = r *W = r*d theta/dt = 2*pi/18.184 = 0.345 = 0.35 m/s

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