# 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?

12.32 rad

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!!

### 3 Answers

- NCSLv 75 months agoBest Answer
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

If you find this helpful, please award Best Answer!

- WhomeLv 75 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

α = 2(8.338) / 1.39(0.119) = 101 rad/s² ANSWER 1

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

- Anonymous5 months ago
The solution you seek is:

F = ma, (0.85 * 0.119)

sin(F/360) * 1.5, Radians and Degrees are identical.