Work and Energy of Rigid Bodies

Introduction

The work-energy principle for rigid bodies relates the work done by external forces and moments to the change in both translational and rotational kinetic energy. Unlike particles, rigid bodies may translate, rotate, or do both simultaneously.

Kinetic Energy

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A rigid body’s kinetic energy is the sum of its translational and rotational kinetic energy:

T = 1 2 m v G 2 + 1 2 I G ω 2 T = \frac{1}{2}mv_{G}^{2} + \frac{1}{2}I_{G}\omega^{2}

If the body rotates about a fixed axis:

T = 1 2 I O ω 2 T = \frac{1}{2}I_{O}\omega^{2}

where I O I_{O} is about the axis of rotation

Work of External Moments

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Unlike particle energy, couple moments can perform work:

U M = ∫ θ 1 θ 2 M ( θ ) d θ U_{M} = \int_{\theta_{1}}^{\theta_{2}}M(\theta)d\theta

If the couple is constant, U M = M θ U_{M} = M\theta

Conservation of Energy

If only conservative forces act, mechanical energy is conserved:

Rolling without slipping:

m g h = 1 2 m v G 2 + 1 2 I G ω 2 mgh = \frac{1}{2}mv_{G}^{2} + \frac{1}{2}I_{G}\omega^{2}

Body rotating about a fixed axis:

m g h + 1 2 I O ω 1 2 = 1 2 I O ω 2 2 mgh + \frac{1}{2}I_{O}\omega_{1}^{2} = \frac{1}{2}I_{O}\omega_{2}^{2}

Spring driving rotation:

1 2 k x 1 2 + 1 2 I ω 1 2 = 1 2 k x 2 2 + 1 2 I ω 2 2 \frac{1}{2}kx_{1}^{2} + \frac{1}{2}I\omega_{1}^{2} = \frac{1}{2}kx_{2}^{2} + \frac{1}{2}I\omega_{2}^{2}

Use conservation of energy only when no nonconservatives forces or applied couples do work. If friction or an applied torque does work, use the principle of work and energy ( T 1 + Σ U 1 → 2 = T 2 T_{1} + \Sigma U_{1 \rightarrow 2} = T_{2} )

Example 1:
Find the distance the 10-kg block must fall in order for the 40-kg spool to have an angular velocity of 15 rad/s. The radius of gyration about center O is 0.3 m.

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I O = m k O 2 = ( 40 k g ) ( 0.3 m ) 2 = 3.6 k g ⋅ m 2 I_{O} = mk_{O}^{2} = (40\ kg)(0.3\ m)^{2} = 3.6\ kg \cdot m^{2}

v = r ω = ( 0.3 m ) ( 15 r a d / s ) = 4.5 m / s v = r\omega = (0.3\ m)(15\ rad/s) = 4.5\ m/s

m g h = 1 2 m v O 2 + 1 2 I O ω 2 mgh = \frac{1}{2}mv_{O}^{2} + \frac{1}{2}I_{O}\omega^{2}

( 10 k g ) ( 9.81 m / s 2 ) ( h ) = 1 2 ( 10 k g ) ( 4.5 m / s ) 2 + 1 2 ( 3.6 k g ⋅ m 2 ) ( 15 r a d / s ) 2 (10\ kg)(9.81\ m/s^{2})(h) = \frac{1}{2}(10\ kg)(4.5\ m/s)^{2} + \frac{1}{2}(3.6\ kg \cdot m^{2})(15\ rad/s)^{2}

Solving for distance: h = 5.16 m h = 5.16\ m

Example 2:
Find the angular velocity of the 10-kg uniform disk and 3-kg uniform slender rod when it has rotated clockwise 90°. It is released from rest.

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90 ∘ = π 2 r a d 90{^\circ} = \frac{\pi}{2}\ rad

Since the moment is constant, U M = M θ = ( 30 N ⋅ m ) ( π 2 r a d ) = 47.124 J U_{M} = M\theta = (30\ N \cdot m)(\frac{\pi}{2}\ rad) = 47.124\ J

The total work done by gravity is the sum of the work of gravity on the rod and disk:

U g = U r + U d = m r g h r + m d g h d = ( 3 k g ) ( 9.81 ) ( 1 m ) + ( 10 k g ) ( 9.81 ) ( 2.4 m ) = 264.8 J U_{g} = U_{r} + U_{d} = m_{r}gh_{r} + m_{d}gh_{d} = (3\ kg)(9.81)(1\ m) + (10\ kg)(9.81)(2.4\ m) = 264.8\ J

Finally, the moment of inertia about A:

I A , r o d = 1 3 m L 2 = 1 3 ( 3 k g ) ( 2 m ) 2 = 4 k g ⋅ m 2 I_{A,\ rod} = \frac{1}{3}mL^{2} = \frac{1}{3}(3\ kg)(2\ m)^{2} = 4\ kg \cdot m^{2}

I A , d i s k = 1 2 m r 2 + m d 2 = 1 2 ( 10 k g ) ( 0.4 m ) 2 + ( 10 k g ) ( 2.4 ) 2 = 58.4 k g ⋅ m 2 I_{A,\ disk} = \frac{1}{2}mr^{2} + md^{2} = \frac{1}{2}(10\ kg)(0.4\ m)^{2} + (10\ kg)(2.4)^{2} = 58.4\ kg \cdot m^{2}

I A = 4 k g ⋅ m 2 + 58.4 k g ⋅ m 2 = 62.4 k g ⋅ m 2 I_{A} = 4\ kg \cdot m^{2} + 58.4\ kg \cdot m^{2} = 62.4\ kg \cdot m^{2}

Principle of work and energy:

T 1 + Σ U 1 → 2 = T 2 T_{1} + \Sigma U_{1 \rightarrow 2} = T_{2}

Since it is released from rest:

0 + U M + U g = 1 2 I A ω 2 0 + U_{M} + U_{g} = \frac{1}{2}I_{A}\omega^{2}

47.124 J + 264.8 J = 1 2 ( 62.4 k g ⋅ m 2 ) ( ω 2 ) 47.124\ J + 264.8\ J = \frac{1}{2}(62.4\ kg \cdot m^{2})(\omega^{2})

ω = 3.16 r a d / s ⟳ \omega = 3.16\ rad/s\ ⟳