Question

A 1.70-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end.If you ignore the weight of the bar itself, how far from the left end of the barbell is the center of gravity? Where is the center of gravity if the 9.00 kg mass of the barbell itself is taken into account?

A 230 g , 24-cm-diameter plastic disk is spun on an axle through its center by an electric motor. What torque must the motor supply to take the disk from 0 to 2000 rpm in 5.0 s ?

Answer 1

Solution: Let m₁ = 25.0 kg be placed at the origin thus its distance from the origin is x₁ = 0 m

m₂ = 37.0 kg is placed at x₂ = 1.70 m

Assuming the mass of the bar to be zero, the center of gravity x_cog can be found by we have

x_cog = (x₁*m₁ + x₂*m₂)/(m₁+m₂)

x_cog = (0m*25kg + 1.70m*37.0 kg)/(25.0kg + 37.0kg)

x_cog = 1.0145 m

Thus the barbell’s center of gravity is at 1.0145 m from the leftend.

Now if we consider the bar’s mass m_b = 9.00 kg. From symmetry we know that the center of mass of the bar is at the midpoint of the bar, that is x_b = (1.70m)/2 = 0.85 m from the left end (or right end). Thus the new center of gravity considering the mass of the bar is,

x_cog = (x₁*m₁ + x₂*m₂ + x_b*m_b)/(m₁+m₂+m_b)

x_cog = (0m*25kg + 1.70m*37.0 kg + 0.85m*9.00kg)/(25.0kg + 37.0kg + 9.00kg)

x_cog = 0.9937 m

Thus the barbell’s new center of gravity in this case is 0.9937 m from the left end.

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Mass of disk m = 230 g = 0.23 kg

Diameter of disk d = 24 cm = 0.24 m

Radius of disk r = 0.12 m

Moment of inertia of disk about central axis I = (1/2)m*r²

I = (1/2)m*r²

I = (1/2)*0.23kg*(0.12m)²

I = 1.656*10^-3 kg.m²

Time of angular acceleration Δt = 5.00 s

Initial angular speed ω_i = 0 rad/s

Final angular speed ω_f

ω_f = 2000 revolutions per minute

ω_f = 2000 rpm*(2π rad/1rev)*(1min/60s)

ω_f = 209.4395 rad/s

The angular acceleration α can be found by

α = (ω_f - ω_i)/Δt

α =[ (209.4395 – 0)rad/s] / 5.0 s

α = 41.8879 rad/s²

Now the torque τ = I*α

τ = (1.656*10^-3 kg.m²)*( 41.8879 rad/s²)

τ = 6.9366*10^-2 N.m