If m1 = 7.50 kg is located at x1 = -8.90 m and m2 = 11.0 kg is located at x2 = 9.9 m, then

Question

If m₁ = 7.50 kg is located at x₁ = -8.90 m and m₂ = 11.0 kg is located at x₂ = 9.9 m, then

A. Where on the x-axis is this system's center of mass?

X_CM =  m

B. How far is each mass from the center of mass?

r₁ =  m

r₂ =  m

C. What is the system's moment of inertia about an axis through the center of mass and perpendicular to the x-axis?

I = kg m²

D. If a constant 43.0 Nm torque acted on this system for 12 seconds, what angular speed would it reach assuming it began from rest?

ω= rad/s

E. calculate the system's kinetic energy and angular momentum using the angular speed reached in part c

K= J

L= kg m²/s

Answer 1

A.

The center of mass is the point x around which the two masses would be balanced.

That means, around which they would have the same torque (but opposite direction).

m₁(x - x₁) = m₂(x₂ - x)

The solution of this equation is

x = (m₁x₁ + m₂x₂)/(m₁ + m₂)

The above is the general formula for calculating center of mass.

You sum up all the masses together with their positions and divide it by

the total mass.

B.

Once you plug in actual numbers in the above equation, you get x,

and from that the two distances.

C.

The answer to this question depends on the shape of the two masses.

If we can assume them to be point masses, then the moment of initial is simply

the sum

I = m₁(x₁-x)² + m₂(x₂-x)²

D.

The formula to use is the analogue of Newton's Second Law:

α = τ/I

where

α is angular acceleration,

τ is torque,

I is moment of inertia.

Once you know α, you can calculate ω from definition of acceleration:

ω = αt

where t is the time the system is subject to the acceleration α.

E.

The formulas to use are

K = Iω²/2

L = Iω