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A steel cylinder of 2m diameter and 0.5m length, with density 7800 kg/m3, rotates at 1500 rev/min. Calculate its volume, mass, and moment of inertia.

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4 parts
a

1.571 m31.571 \, \text{m}^3

b

12252 kg12252 \, \text{kg}

c

6126 kg⋅m26126 \, \text{kg} \cdot \text{m}^2

d

157.1 rad/s157.1 \, \text{rad/s}

Step-by-step solution

5 steps

Let's solve Question 7, taking into account the dimensions provided in Figure 7.

Part 7

(10 marks)

A full steel cylinder of 2000 mm2000 \, \text{mm} diameter is turning about a fixed axis (GZ) with the rotational velocity of 1500 rev/min1500 \, \text{rev/min}. The volume density of the steel is ρ=7800 kg/m3\rho = 7800 \, \text{kg/m}^3. Recall: JGZ=MR22J_{GZ} = \frac{MR^2}{2}.

From Figure 7:
Diameter D=2000 mm=2.000 mD = 2000 \, \text{mm} = 2.000 \, \text{m}
Length L=500 mm=0.500 mL = 500 \, \text{mm} = 0.500 \, \text{m}

Part a

1.571 m31.571 \, \text{m}^3

Determine the volume VV of the cylinder.

  1. Calculate the radius and then the volume of the cylinder.

    R=D2=2.000 m2=1.000 mR = \frac{D}{2} = \frac{2.000 \, \text{m}}{2} = 1.000 \, \text{m} V=πR2LV = \pi R^2 L V=π(1.000 m)2(0.500 m)V = \pi (1.000 \, \text{m})^2 (0.500 \, \text{m}) V=0.5π m3V = 0.5\pi \, \text{m}^3 V≈1.571 m3V \approx \boxed{1.571 \, \text{m}^3}

Part b

12252 kg12252 \, \text{kg}

Calculate the mass MM of the cylinder.

  1. Calculate the mass of the cylinder using its volume and the given density.

    M=ρVM = \rho V M=(7800 kg/m3)(0.5π m3)M = (7800 \, \text{kg/m}^3)(0.5\pi \, \text{m}^3) M=3900π kgM = 3900\pi \, \text{kg} M≈12252 kgM \approx \boxed{12252 \, \text{kg}}

Part c

6126 kg⋅m26126 \, \text{kg} \cdot \text{m}^2

Determine the moment of inertia JGZJ_{GZ} of the cylinder.

  1. Calculate the moment of inertia using the given formula, mass, and radius.

    JGZ=MR22J_{GZ} = \frac{MR^2}{2} JGZ=(3900π kg)(1.000 m)22J_{GZ} = \frac{(3900\pi \, \text{kg})(1.000 \, \text{m})^2}{2} JGZ=1950π kg⋅m2J_{GZ} = 1950\pi \, \text{kg} \cdot \text{m}^2 JGZ≈6126 kg⋅m2J_{GZ} \approx \boxed{6126 \, \text{kg} \cdot \text{m}^2}

Part d

157.1 rad/s157.1 \, \text{rad/s}

Express the angular velocity ω\omega.

  1. Convert the rotational velocity from revolutions per minute to radians per second.

    ω=1500 rev/min×2π rad1 rev×1 min60 s\omega = 1500 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} ω=1500×2π60 rad/s\omega = \frac{1500 \times 2\pi}{60} \, \text{rad/s} ω=50π rad/s\omega = 50\pi \, \text{rad/s} ω≈157.1 rad/s\omega \approx \boxed{157.1 \, \text{rad/s}}

Part e· 1 step

Determine the kinetic energy KEK_E.

  1. Calculate the rotational kinetic energy using the moment of inertia and angular velocity.

    KE=12JGZω2K_E = \frac{1}{2} J_{GZ} \omega^2 KE=12(1950π kg⋅m2)(50π rad/s)2K_E = \frac{1}{2} (1950\pi \, \text{kg} \cdot \text{m}^2) (50\pi \, \text{rad/s})^2 KE=12(1950π)(2500π2) JK_E = \frac{1}{2} (1950\pi) (2500\pi^2) \, \text{J}
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