A sphere of radius 20 cm and mass 1.2 kg rolls without slipping down a 30 degree incline that is 10 m long.
a) Calculare the translational and rotational speeds of the sphere when it reaches the bottom. Assume it starts from rest, and first solve for each speed symbolically, and the numerically.
b)What is the ratio of translational to rotational KE (KE/KErot) at the bottom? Also solve this ratio symbolically and then numerically so you can answer the following question.
C) Do your answers in parts a or b depend on the radius of the sphere or its mass.
My teacher did part a) for symbolically solving for the speed, and ended up with: w= 1/r*sqrt 10/7 gh
He began with mgh= 1/2mv^2 = 1/2(2/5mv)(v/r)^2
I don't know where to go from there.
To solve part a) symbolically for the translational and rotational speeds, we can use the principle of conservation of energy.
Step 1: Calculate the gravitational potential energy (PE) of the sphere at the top of the incline.
Given: mass (m) = 1.2 kg, height (h) = 10 m, g = acceleration due to gravity = 9.8 m/s^2.
PE = mgh = 1.2 kg × 9.8 m/s^2 × 10 m = 117.6 J
Step 2: Calculate the total energy (E) at the top of the incline.
The total energy is the sum of potential energy and kinetic energy.
E = PE = 117.6 J
Step 3: Calculate the translational kinetic energy (KEtrans) at the bottom of the incline using the total energy at the top.
KEtrans = E - PE = 117.6 J - 0 J = 117.6 J
Step 4: Solve for the translational velocity (v) using the formula for translational kinetic energy.
KEtrans = 1/2 mv^2
117.6 J = (1/2) × 1.2 kg × v^2
v^2 = (117.6 J × 2) / (1.2 kg)
v^2 = 196 m^2/s^2
v = √(196 m^2/s^2)
v = 14 m/s (approximately)
Step 5: Calculate the rotational kinetic energy (KErot) at the bottom of the incline using the total energy at the top.
KErot = E - KEtrans
KErot = 117.6 J - 117.6 J = 0 J
Step 6: Solve for the rotational velocity (ω) using the formula for rotational kinetic energy.
KErot = (1/2) I ω^2
0 J = (1/2) × (2/5) m r^2 × ω^2
Simplifying, we have:
0 = (1/5) m r^2 × ω^2
Since both rotational kinetic energy (KErot) and the radius (r) are non-zero, the only way this equation can hold true is if ω = 0. Therefore, at the bottom of the incline, the sphere has no rotational speed.
To solve for the speeds numerically, you would substitute the given values into the equations and calculate the results.
In summary, the translational speed (v) is approximately 14 m/s, and the rotational speed (ω) is 0 rad/s, symbolically and numerically.