# Physics homework #8

1.Part 1: A cylinder w/moment of inertia 30.9 kg x meters2 rotates w/angular velocity 8.48 rad/sec on a frictionless vertical axle. A second cylinder, w/moment of inertia 27.3 kg Xmeters2, initially not rotating, drops onto the first cylinder and remains in contact. Since the surfaces are rough, the two eventually reach the same angular velocity. Calculate the final angular velocity. Answer in units of rad/sec.  Part 2: Show that energy is lost in the above situation by calculating the ratio of the final to the initial kinetic energy.

2. Part 1: A uniform rod pivoted at one end (Point O) is free to swing in a vertical plane in a gravitational field, held in equilibrium by a force F at its other end. The rod of length 7m and weight of 6.3N makes an angle 40 degrees with the horizontal and the force an angle of 60 degrees w/the horizontal. What is the magnitude of the force F? Answer in units of N. Part 2: What is the magnitude of the force the pivot exerts on the rod at point O?

3. Part 1: An airplane of mass 17991 kg flies level to the ground at an altitude of 10km w/a constant speed of 162m/s relative to the Earth. What is the magnitude of the airplane’s angular momentum relative to a ground observer directly below the airplane in kg . m2 /sec? Answer in units of kg . m2/s. Part 2: Does this value change as the airplane continues its motion along a straight line?

1. Yes. L changes w/certain period as the airplane moves.

2. Yes L decreases as the airplane moves.

3. No. L = constant.

4. Yes. L increases as the airplane moves.

4. At t = 0, a wheel rotating about a fixed axis at a constant angular deceleration of 0.56 rad/sec2 has an angular velocity of 2.5 rad/sec and an angular position of 6.2 rad. What is the angular position of the wheel after 3 sec? Answer in units of rad.

1.

There is no external applied force hence no external momentum of force. The total kinetic moment ($J=I*\omega$) is the same before and after the second cylinder is dropped. The situation is similar to a plastic collision of two bodies where the total linear momentum is the same before and after the collision. Before the second cylinder is dropped.

$J_1 =I_1*\omega_1 = 30.9*8.48 =262.032 kg*m^2/s^2$

After the second cylinder is dropped

$J_2 = (I_1+I_2)*\omega_2 =(30.9+27.3)*\omega_2 =58.2*\omega_2$

$J_1=J_2$

$\omega_2 =262.032/58.2 =4.5 (rad/sec)$

The kinetic energy before the second cylinder is dropped is

$Ec_1 =I_1*\omega_1^2/2 =30.9*8.48^2/2 =1159.23 J$

and after the second cylinder is dropped

$Ec_2 =(I_1+I_2)*\omega_2^2/2 =58.2*4.5^2/2 =589.275 J$

$Ec_2/Ec_1 =589.275/1159.23 =0.508 <1$ hence the kinetic energy is lost.

2.

The rod is in equilibrium. The moment of the force F with respect to point O is equal to the moment of the rod weight with respect to the same point.

Let l be the length of the rod and $F_n$ the vertical force applied at the free end of the rod.

let alpha be the angle the rod is making with the horizontal and beta the force is making with the horizontal. The weight is applied at half the rod distance

$F_n*l*cos(\alpha) = G*l/2 *cos(\alpha)$

$F*sin(\beta)*l =G*l/2$

$F =G/(2*sin(\beta)) =6.3/2/sin(60) =3.637 N$

Part 2. The reaction R in point O is as follows.

The vertical reaction is $R_v =G-F_n =6.3-3.637 =2.663 N$

The horizontal reaction $R_h = F_x = F*cos(\beta) =3.637*cos(60) =1.8185 N$

The total reaction at point O is $R = \sqrt{R_v^2+R_h^2} =\sqrt{2.663*2.663+1.8185*1.8185} =3.225 N$

3.

1. The angular momentum (or momentum of momentum) of a body with respect to a point O is by DEFINITION

L =r x p where r is the distance of the body from the point O and $p=m*v$ the linear momentum of the body, and x is the vector product

Hence for an observer directly  below the plane

$L =r*m*v =10000*17991*162 =2914542*10^4 kg*m^2/s$

2. The variation of the angular momentum is equal to the torque of the external force applied. There is no external force applied hence no variation of angular momentum as the plane moves. The correct answer is 3.  No. L= constant.

4.

The equation of uniform accelerated motion is

$s =s_0+ v_0*t+ a*t^2/2$ where s is the position, $v_0$ is the initial velocity, $a$ the acceleration and $t$ the time.

Analog for rotation

$\alpha= \alpha_0 +\omega_0*t + \epsilon*t^2/2$

where $\alpha$ is the angular position, $\omega_0$ the initial angular speed, $\epsilon$ the angular acceleration and $t$ the time

$\alpha =6.2 +2.5*3 -0.56*3*3/2 =11.18 rad$