The system shown consists of a rotating base a 10kg ball wit

The system shown consists of a rotating base, a 10kg ball with a radius of 0.5m and a connecting link. The base is rotating about the +z-axis with a constant angular speed as shown. Meanwhile, the link is rotating about a pin joint connected to the base (aligned with the x-axis at the instant shown) with the constant magnitude and sense shown. If the link is 5m long from the center of the pivot to the center of the ball, find parts a-e at the instant shown. Neglect the mass of everything but the ball. Report your answers as vectors in terms of the shown coordinate system. a) The linear momentum of the ball b) The angular momentum of the ball about the origin of the xyz coordinate system c) The kinetic energy of the ball d) The forces developed at the connection between the ball and the link e) The reaction moments developed at the connection between the ball and the link

Solution

Here the ball and the pin is rotating with respect to the base which itself is rotating about the z axis with = 1 rad/sec

The ball is spherical in shape wich radius R = 0.5 m and the distance from the origin here is L = 5 m.

Part a.) The ball, at the given instant, will have the net velocity as a result of two both the revolving motion.

The linear velocity due to the rotation of the link would be given as:V_L = ₂(L) = 10 m/s in a direction perpendicular to the link.

The linear velocity due to the rotation of the base would be given as:Vb = ₁(4/5)5 = 4 m/s into the page, that is, along the x axis.

Now apparently, V_L and Vb are both perpnendicular to each other, hence the net velocity would be given as:

Vnet = sqrt(16 + 100) = 10.77 m/s

Therefore the linear momentum of the ball would be = mv = 10(10.77) = 107.7 kg-m/s

Part b.) As above, the ball will have two angular velocities which again would be perpendicular to each other.

Hence the net angular velocity about the origin would be given as: _net = sqrt[ ₁² + ₂²] = sqrt(4 + 1) = 2.236 rad/sec

Also, the moment of inertia of the ball about the origin would be given as: (2/5)Mr² + ML²

That is, I = 0.8 x 10 x 0.25 + 10 x 25 = 252 kg-m²

Therefore the net angular momentum about the origin would be: I = 252 (2.236) = 563.472

Part c.) The kinetic energy of the ball would be given as: 0.5I² = 0.5 x 252 x 2.236² = 629.96 J

Part d.) While rotating along a circular path, we know that the net force, that is the centripetal force, acting towards thec centre is given as: F = m²R

Now, as the ball will have following forces acting on it: a force along - Y axis due to rotation of the base about the axis, weight of the ball downwards and centripetal force along the direction of the link due to rotation of the link

So, we have force along the Y axis = -m₂²L(0.8) - m₁²L(0.8) = - 4 x 10 x [4 + 1] = -200 N

The force along the Z axis = mg = 100 N

Therefore the force acting on the ball due to the link would be given as: F = -200 j + 100 k

Part e.) The reaction moment of the forces at the developed between the link and the ball would be given as the cross product of the force and the distance vector.

That is, Moment due to the force along the Y axis about the origin would be: 200(3) = 600 N-m

Moment due to the force along the Z axis about the origin would be: 100(3) = 300 N-m