Anonymous
Anonymous asked in Science & MathematicsPhysics · 7 months ago

# Physics help?

M, a solid cylinder (M=1.39 kg, R=0.119 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.850 kg mass, i.e., F = 8.338 N. Calculate the angular acceleration of the cylinder.

If instead of the force F an actual mass m = 0.850 kg is hung from the string, find the angular acceleration of the cylinder.

How far does m travel downward between 0.510 s and 0.710 s after the motion begins?

The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.505 m in a time of 0.470 s. Find Icm of the new cylinder.

I was able to answer the first two questions but am struggling with the last two. Thanks so much!!

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• NCS
Lv 7
7 months ago

You answered the first two, so you have the angular acceleration in the second case. Let's call it α.

Then the linear acceleration of the mass is

a = α*r = α * 0.119m

and the distance the mass travels between the two times is

s = ½ * a * [(0.710s)² - (0.510s)²] = a * 0.122s²

OR

if the "12.32 rads" is correct for the angular displacement between the two times,

s = Θ*r = 12.32rad * 0.119m = 1.47 m

For the last part, just work it backwards:

s = ½at²

0.505 m = ½ * a * (0.470s)²

a = 4.57 m/s²

α = a/R = 38.4 rad/s²

the tension in the string both restrains the falling mass and torques the cylinder:

T = m(g - a) = 1.39kg * (9.8 - 4.57)m/s² = 7.27 N

torque τ = I*α = T*r

I * 38.4rad/s² = 7.27N * 0.119m

I = 0.0225 kg·m²

• Whome
Lv 7
7 months ago

the basic concept is torque τ will equal moment of inertia I multiplied by angular acceleration α

τ = Iα

α = τ/I

The moment of inertia for a uniform cylinder is I = ½MR²

The torque is a force acting at a distance τ = FR

α = FR / ½MR² = 2F / MR

When you make the torque force due to an actual mass m, that mass must also accelerate. In respect to the pivot point, that mass will appear as a point mass on the radius of the cylinder

α = FR / (½MR² + mR²) = mgR / R² (m + ½M) = mg / R(m + ½M)

α = 0.850(9.81) / 0.119(0.850 + ½(1.39)) = 45.4 rad/s² ANSWER 2

How far does m travel downward between 0.510 s and 0.710 s after the motion begins?

θ = ½α(tf²) - ½α(ti²) = ½(45.4)(0.710² - 0.510²) = 5.53 radians

d = Rθ = 0.119(5.53) = 0.658 m ANSWER 3

The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.505 m in a time of 0.470 s. Find Icm of the new cylinder.

θ = d/R = 0.505 / 0119 = 4.24 radians

θ = ½αt²

α = 2θ / t² = 2(4.24) / 0.470² = 38.4 rad/s²

α = mgR / (Icm + mR²)

(Icm + mR²) = mgR / α

Icm = (mgR / α) - mR²

Icm = mR(g/α - R)

Icm = 0.850(0.119)((9.81 / 38.4) - 0.119)

Icm = 0.0138 kg•m² ANSWER 4