Dynamics Problem Solutions
Problem D1
Use the following equation:
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Substitute g=32.2, y0=200, v0y=400sin(30) and y=0. Solve the resulting equation for t. The result is t=13.3 seconds. Then use the following equation:
and find the distance covered. The answer is: x=4626 feet.
Problem D2
Find an as a function of time using:
and ,
Where at=7 and r=300. Substitute into the expression for the magnitude of the acceleration:
The magnitude of acceleration is given as g/4. Solve for t. The answer is t=4.93 sec. And the velocity is v=34.5 ft/sec. The velocity is not asked for in the problem statement.
Problem D3
Using find given r=9 and =10. Using given , find the tangential acceleration. Finally find the magnitude of the acceleration vector:. The answer is v=7.5 ft/sec, and =80.8 ft/s2.
Problem D4
Use the polar coordinates formulation: and and find the magnitude of acceleration vector by combining the two using:
The important point is that the two components of the tangential acceleration point in opposite directions and must be subtracted (refer to lecture notes for helpful graphics). Substitute for:
The answer is a=242 ft/sec2
Problem D5
The ball’s normal acceleration is which evaluates to 84.5 ft/s2 using v=13 ft/sec and r=2 ft. The equation of motion in the radial (normal) direction is:
where g=32.2 and W=25 lbs. We can solve for T. The answer is T=87.25 lbs. The equation of motion in tangential direction is:
.
We can solve for the tangential acceleration and find the angular acceleration using: . The answer is a=16.1 ft/s2 and =8.05 rad/s2. The problem asks for the angular acceleration but the answer given is a – so accept either.
Problem D6
Using conservation of energy: . In this formula, x is the difference between the stretched length and the free length of the spring and is calculated using the Pythagorean formula: . The change is kinetic energy is calculated to be 0.087v2. The change is potential energy is calculated to be -5.244. We can solve for v. The answer is v=8.2 ft/s.
Problem D7
The two positions for writing the energy equation is the point of release and the position where the block comes to a rest. The change in kinetic energy is
.
The change in gravitational P.E. is
where is the maximum amount the spring gets compressed. The change in elastic (spring) PE is:
There is also work done on the particle by friction:
Where f is the friction force, is the coefficient of friction, and N is the normal force equal to Wcos().This work is always negative. Now, using the energy relationship:
We can find . The answer is=1.637 in and Fmax=163.7 lbs.
Problem D8 :
The change in KE as well the change in PE are the same as in the notes.
The work done by the external force on the roller is U=Fd=(-1)(3)=-3 ft-lbs. Therefore:
The answer is w=3.2 rad/s
Problem D9
The change in K.E. is:
.
The pulley moment of inertia can be found from its radius of gyration and its mass using:
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The velocity of the pulley center is related to the pulley’s angular velocity: . Therefore the change in K.E. is a function of .
The change in gravitational P.E. is
The change in spring potential energy is:
.
Where, x1 and x2 are the initial and final length differences with respect to the free length. When the block falls 9 inches, the combined length of the rope and the spring need to increase 18 inches which is all due to spring stretching as the rope remains the same length. Therefore: x1=0, and x2=18.
Settingup the energy equation
we can find =12.12 rad/s and v=4.04 ft/s.
Problem #D10
By running the physical event in reverse, we can find the bullet’s velocity. From the energy relation and the swing of the pendulum, we can find the bar’s angular velocity just after impact. Then, from the angular momentum relationship before and after impact we can find the bullet’s speed.
Substituting into the energy equation and solving for , we get i=8.55 rad/sec. This is the angular velocity of the pendulum after impact. From angular momentum relationship before and after impact:
Equating angular momentums and solving for v we get: v=226 ft/sec.