inelastic collisionkinetic energy loss

Mid Term 3

Multiple Choice

An object is projected vertically upward from the surface of a nonrotating planet of radius RR, with an initial velocity equal to 48% of the escape velocity for the planet. THe maximum distance from the center of the planet attained by the object is closed to

Solution

1.3R1.3R

(1) A wheel has a radius of 0.40m and is mounted on frictionless bearing. A block is suspended from a rope which is wound on the wheel and the attached to it. The wheel is released from rest and the block descends 1.5m in 2.00s. The tension in the rope during the descent of the block is 20N. In the firgure, the moment of inertia of the wheel is closest to

Solution

4.3 kgm24.3\text{ kg} \cdot \text{m}^2

(2) The radius of a 3.0kg wheel is 6.0cm. The wheel is released from rest at point A on a 30°30\degree incline. The wheel rolls without slipping and moves 2.4m to point B in 1.20s. The angular acceleration of the wheel is closest to:

Solution

56 rad/s256 \text{ rad}/\text{s}^2

(3) A spining ice skater is able to contronl the rate at which she rotates by pulling in her arms. We can best understand this effect by observing that in this process.

Solution

Her angular momentum remains constant.

(4) When is the angular momentum of a system constant?

Solution

When no torque acts on the system.

(5) A meteor of mass about 3.4×1012kg3.4 \times 10^{12}kg (a chunk of ice about a mile in diameter) is heading straight for Jupiter. When it hits, there will be a huge release of energy, visible here on earth. Assuming it has fallen from far away, how much energy will be released when it hits Jupiter? The radius of Jupiter is about 7×107m7 \times 10^7 m and its mass is 1.9×1027kg1.9 \times 10^{27}kg.

Solution

6.1×1021J6.1 \times 10^{21}J

(6) Water (density =1.0×103kg/m3=1.0 \times 10^3 kg/m^3) flows through a horizontal tapered pipe. At the wide end its speed is 4.0m/s4.0m/s. The difference in pressure between two ends is 4.5×103Pa4.5 \times 10^3 Pa. The speed of the water at the narrow end is:

Solution

5.0m/s5.0m/s.

(10) At time t=0t=0, a wheel has an initial angular velocity ω0=2\omega_0 = 2 rad/s. The wheel is experiencing an angular acceleration α=3t\alpha = 3t for 3 seconds. Find the final angular velocity.

Solution

ωf=ω0+033tdt=2.0+32t203=2.0+(13.50)=15.5rad/s\begin{aligned} \omega_f &= \omega_0 + \int_0^3 3t dt\\ &= 2.0 + \frac{3}{2}t^2\mid_0^3\\ &= 2.0 + (13.5 - 0) = 15.5 rad/s \end{aligned}

Applications

(2) A constant tangential force F=500N acts on a rotating disk of mass M=50kg and radius=5m. Four 50kg kids stand at the outer edge of the rotating disk(at R=5m). (a) Calculate angular acceleration α\alpha of the system. (b) Calcuate WfW_f is the force acts for 5sec. (c) Calculate WW if the force is stopped , and all the kids run to the center.