As you approach the I-90 exit in your uncle's little red sports car convertible, you happen to notice a cat crossing the road. Not that you have any particular love of cats (especially this one, whose snooty disposition seems to beckon for a game of chicken), but you swerve out of the way, spilling scalding hot coffee all over your new summer duds.
Unknown to you at the time, the auto mechanic, having had a late night out and consequently being somewhat in a fog, forgot to fasten the lug nuts on one of your uncle's tires.
Given your velocity on the exit ramp as 3.2 m/s, a tire mass of 12 kg and radius 29 cm;
a) What is the angular momentum L of uncle's tire immediately upon its departure from its axle (assume the tire is a solid cylinder) in kg m2/s?
I solved this to get 5.57 kg m2/s.
b) Next the tire cruises up an embankment at an angle θ = 27° above the horizontal for 1 meter before launching across Piranha Lake. What is the velocity vfinal of the tire at the top of the embankment in m/s?
For this part I'm having trouble expressing my total kinetic energy (that I solved for) in terms of velocity.
I plugged in the numbers for the initial KE: (1/2)*I*ω2 + (1/2)*m*v2. where I got 92.16 J.
Final KE = Initial KE - m*g*h where h = 1m * sinθ.
My final KE came out to be 38.7707 J.
How do I use this to find the final velocity?
To find the final velocity of the tire at the top of the embankment, you can use conservation of energy.
The initial kinetic energy of the tire is equal to the sum of its rotational kinetic energy and its translational kinetic energy. The rotational kinetic energy is given by (1/2) * I * ω^2, where I is the moment of inertia of the tire and ω is the angular velocity. The translational kinetic energy is given by (1/2) * m * v^2, where m is the mass of the tire and v is its linear velocity.
The final kinetic energy of the tire is equal to the initial kinetic energy minus the work done against gravity as it moves up the embankment. The work done against gravity is given by m * g * h, where g is the acceleration due to gravity and h is the height the tire moves up.
Setting these two equations equal to each other, we have:
(1/2) * I * ω^2 + (1/2) * m * v^2 = (1/2) * I * ω^2 + (1/2) * m * v^2 - m * g * h
Simplifying the equation, we have:
m * g * h = 0
Since h is equal to 1m * sin(θ), and θ is given as 27°, we can substitute the values:
m * g * (1m * sin(27°)) = 0
Solving for g, we have:
g = 9.8 m/s^2 (acceleration due to gravity)
Substituting this value, we have:
m * (9.8 m/s^2) * (1m * sin(27°)) = 0
Simplifying further, we have:
m * (9.8 m/s^2) * sin(27°) = 0
Now, we can solve for the final velocity by rearranging the equation:
(1/2) * I * ω^2 + (1/2) * m * v^2 = (1/2) * I * ω^2 + (1/2) * m * v^2 - m * g * h
(1/2) * m * v^2 = m * g * h
Multiply both sides of the equation by 2:
m * v^2 = 2 * m * g * h
Divide both sides of the equation by m:
v^2 = 2 * g * h
Now substitute the known values:
v^2 = 2 * (9.8 m/s^2) * (1m * sin(27°))
Simplifying further, we have:
v^2 = 2 * 9.8 m/s^2 * 0.454 m
v^2 = 8.87 m^2/s^2
Taking the square root of both sides, we have:
v = √(8.87) m/s
Approximately, the final velocity of the tire at the top of the embankment is:
v ≈ 2.977 m/s
To find the final velocity of the tire at the top of the embankment, you can equate the final kinetic energy to the initial kinetic energy of the tire.
The final kinetic energy is given as:
KE_final = 1/2 * I * ω^2 + 1/2 * m * v^2
Here, I is the moment of inertia of the tire (which you calculated as 5.57 kg m^2/s in part a), ω is the angular velocity of the tire (which can be calculated using the equation ω = v/r, where v is the linear velocity and r is the radius of the tire), m is the mass of the tire (given as 12 kg), and v is the final linear velocity of the tire.
To express the final kinetic energy in terms of velocity, you need to substitute the value of angular velocity ω. Rearrange the equation ω = v/r to solve for ω and substitute it into the final kinetic energy equation:
KE_final = 1/2 * (m * v^2 / r^2) * r^2 + 1/2 * m * v^2
KE_final = 1/2 * m * v^2 + 1/2 * m * v^2
KE_final = m * v^2
Now, equate the final kinetic energy (38.7707 J) to m * v^2 and solve for v:
38.7707 J = 12 kg * v^2
v^2 = 38.7707 J / 12 kg
v^2 = 3.2309 m^2/s^2
Taking the square root of both sides:
v = √(3.2309 m^2/s^2)
v ≈ 1.8 m/s
Therefore, the final velocity of the tire at the top of the embankment is approximately 1.8 m/s.