Answer:
a) 236 J b) 30 rev
Step-by-step explanation:
a) As friction force does not work (because it doesn't cause any displacement), this means that the mechanical energy of the cylinder must keep constant, so we can say:
ΔK + ΔU = 0
The change in the gravitational potential energy, taking as zero reference the point in height corresponding to the one 5 m below down the incline, is just as follows:
ΔU = 0 - m*g*h
As the angle of the incline is 37º, applying the definition of sine, we have:
ΔU = -m*g*d*sinθ = -12kg*9.8 m/s²*5m*0.602 = -354 J
The change in kinetic energy will be the sum of the change in translational kinetic energy, plus the change in rotational kinetic energy:
ΔK = 1/2*m* (vf²-v₀²) + 1/2* I* (ωf²-ω₀²)
For a solid cylinder, rotating regarding an axis along its central axis, the moment of inertia I is just m*R²/2.
As the cylinder is rolling without slipping, there exists a fixed relationship between angular and linear velocity, as follows:
ω = v/r
Replacing these two knowns in the equation for ΔK:
ΔK = ΔKt + Δkrot = 1/2 *m* (vf²-v₀²) + 1/2*(m*r²/2)*(vf²-v₀²)/r²
ΔK = m*(vf²-v₀²) /2 + m*(vf²-v₀²) / 4 = m*g*h = 354 J
We can solve for (vf²-v₀²), as follows:
(vf²-v₀²) = 4/36 * 354 J = 354 J / 9 = 39.3 (m/s)²
The change in the translational kinetic energy can be expressed as follows;
ΔKtrans = 1/2*m*(vf²-v₀²) = 1/2*12 Kg* 39.3 (m/s)² = 236J
b) If the angular acceleration is constant, we can use any of the angular equivalents of the kinematic equations.
So, we can write the following equation:
ωf² - ω₀² = 2*Δθ*γ
As the disk started from rest, ω₀ = 0.
Also, we need the Δθ in radians, so we need to convert 10 rev to radians, as follows:
10 rev* (2*π) rad/rev =20*π rad
Replacing in (1), we can solve for ωf²:
ωf² = 2*20*π*γ = 40*π*γ rad²/sec²
Now, we know that keeping constant the angular acceleration , we want to reach to a new value of ω, starting from the value of ωf that we have just found.
⇒ ω = 2*ωf
Using the same equation, we can say:
4*ωf² - ωf² = 2*Δθ*γ
⇒3*ωf² = 2*Δθ*γ
Replacing by the value of ωf we found above:
3*40*πγ = 2*Δθ*γ
As the angular acceleration γ is on both sides, we can factor them out, and solve for Δθ, as follows:
Δθ = 60*π rad
As we need the result in revolutions, we need to convert the radians to revs, as follows:
Δθ = 60*π rad*(1 rev/2*π rad) =30 revs