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Wize University Dynamics Textbook (Master) > Energy Methods for Rigid Bodies

Conservation of Energy for a Rigid Body

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Conservation of Energy for a Rigid Body:


We can continue to use the same general equation as used for particle equilibrium as shown below, with 2 additional terms to consider:

T1+ΣU1−2=T2T_1 + \Sigma U_{1-2}=T_2

But now we must also account for the rotational kinetic energy of a body using:

T=12mvG2+12IGω2T = \frac{1}{2}mv_G^2+\frac{1}{2}I_G\omega^2

And if an external couple moment is applied on the body, then we account for it by:

U1−2=∫θ1θ2MdθU_{1-2} = \int_{\theta_1}^{\theta_2}Md\theta
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The 10 lb double pulley shown below is initially travelling at 20 rad/s clockwise while supporting loads at A and B, initially at the same height. Assuming neither cable slips, and that the weight of blocks A and B are 1 and 5 lb respectively, determine the height difference between the blocks when both blocks come to rest.


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Initially : Block B @ C
Cable Equation :

2xB+xA⇒ 2ΔxB=−ΔxA2x_B +x_A\Rightarrow\ 2\Delta x_B=-\Delta x_A

2VB=−VA2V_B=-V_A
a) When ΔxB=2 ft  ⟹ ΔxA=4 ft \Delta x_B=2\ ft\ \ \Longrightarrow\ \Delta x_A=4\ ft\
set height reference @ xAo

then : T1 = 0
V1 = 0

then : V2=12(632.2)(−4)=0.3727V_2=\frac{1}{2}\left(\frac{6}{32.2}\right)\left(-4\right)=0.3727

T2= 0.3727 = 12((632.2)(2VB)2+(1832.2)(VB)2)T_2=\ 0.3727\ =\ \frac{1}{2}\left(\left(\frac{6}{32.2}\right)\left(2V_B\right)^2+\left(\frac{18}{32.2}\right)\left(V_B\right)^2\right)

VB=0.756 AS→V_B=0.756\ \frac{A}{S}\rightarrow

VA=1.512 AS↓  V_A=1.512\ \frac{A}{S}\downarrow\ \

b) T1+μ1−2=T2 0T_1+\mu_{1-2}=\cancel{T_2}\space^{0}

T1 = 0.3727

(μ1−2)g=0.3727\left(\mu_{1-2}\right)_g=0.3727

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The 10 lb double pulley shown below is initially travelling at 20 rad/s clockwise while supporting loads at A and B, initially at the same height. Assuming neither cable slips, and that the weight of blocks A and B are 1 and 5 lb respectively, determine the height difference between the blocks when both blocks come to rest.


PAGE BREAK
Initially : Block B @ C
Cable Equation :

2xB+xA⇒ 2ΔxB=−ΔxA2x_B +x_A\Rightarrow\ 2\Delta x_B=-\Delta x_A

2VB=−VA2V_B=-V_A
a) When ΔxB=2 ft  ⟹ ΔxA=4 ft \Delta x_B=2\ ft\ \ \Longrightarrow\ \Delta x_A=4\ ft\
set height reference @ xAo

then : T1 = 0
V1 = 0

then : V2=12(632.2)(−4)=0.3727V_2=\frac{1}{2}\left(\frac{6}{32.2}\right)\left(-4\right)=0.3727

T2= 0.3727 = 12((632.2)(2VB)2+(1832.2)(VB)2)T_2=\ 0.3727\ =\ \frac{1}{2}\left(\left(\frac{6}{32.2}\right)\left(2V_B\right)^2+\left(\frac{18}{32.2}\right)\left(V_B\right)^2\right)

VB=0.756 AS→V_B=0.756\ \frac{A}{S}\rightarrow

VA=1.512 AS↓  V_A=1.512\ \frac{A}{S}\downarrow\ \

b) T1+μ1−2=T2 0T_1+\mu_{1-2}=\cancel{T_2}\space^{0}

T1 = 0.3727

(μ1−2)g=0.3727\left(\mu_{1-2}\right)_g=0.3727

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A 20 lb ladder is resting on a smooth wall as shown below, at an angle of 30°. Determine the velocity of A as it hits the ground if the surface at B is smooth. How does your answer change if the coefficient of static and kinetic friction at B were 0.15 and 0.10 respectively? You can treat the ladder as a slender rod.


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checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.
A 20 lb ladder is resting on a smooth wall as shown below, at an angle of 30°. Determine the velocity of A as it hits the ground if the surface at B is smooth. How does your answer change if the coefficient of static and kinetic friction at B were 0.15 and 0.10 respectively? You can treat the ladder as a slender rod.


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