Question

A grinding wheel is a uniform cylinder with a radius of 7.80 cm and a mass of 0.550 kg.

Part A
Calculate its moment of inertia about its center. Express your answer to three significant figures and include the appropriate units.
Part B
Calculate the applied torque needed to accelerate it from rest to 1750 rpm in 7.40 s .

Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 58.0 s .

Answer 1

Answer:

a. I = 167.31 x 10 ⁻³ kg*m²

b. T = 4.59 kg * m² / s²

Explanation:

The moment of inertia of a uniform cylinder:

a.

r = 7.8 cm * 1 m / 100 cm = 0.078 m

I = ½ * m * r²  

I = ½ * 0.55 kg * (0.078²m)

I = 167.31 x 10 ⁻³ kg*m²

b.

T = Iα’ + Iα,    

α’ = ω’/t = 1750 rpm * (2π/60) / 7.40s  =  24.76 rad/s²

α = ω/t = 1500 rpm * (2π/60) / 58  = 2.71 rad/s²

T = (167.31 x 10⁻³ kg*m²)* (24.76 + 2.71 ) rad / s²  

T = 4.59 kg * m² / s²

Answer 2

The moment of inertia and the applied torque of the uniform cylinder calculated using the given parameters are :

• 167.3 x 10 ⁻³ kgm²
• 4.59 kgm²/s²

Given the Parameters :

• Radius, r = 7.8cm = 0.078 m
• Mass , m = 0.55kg
• Time, t = 7.40 s
• Angular velocity, ω = 1750 rpm

Moment of inertia about it's center :

• Moment of Inertia, I = 0.5mr²

I = 0.5(0.55)(0.078²)

I = 167.3 x 10 ⁻³ kgm²

b.)

• Applied torque = T = I(α’ + α)

Converting the velocity to radian per seconds :

α’ = 1750 rpm = (2π/60) / 7.40s = 24.76 rad/s²

α = 1500 rpm * (2π/60) / 58 = 2.71 rad/s²

T = 167.31 x 10⁻³(24.76 + 2.71 )

T = 4.59 kgm²/s²

Therefore, applied torque ls 4.59 kgm²/s²

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