- #1
Eclair_de_XII
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Homework Statement
"Figure 8-32 shows a pendulum of length ##L=\frac{5}{4}m##. Its bob (which effectively has all the mass) has speed ##v_0## when the cord makes an angle ##\theta_0=40°## with the vertical. (a) What is the speed of the bob when it is in its lowest position if ##v_0=8\frac{m}{s}##? What is the least value that ##v_0## can have if the pendulum is to swing down and then up (b) to a horizontal position, and (c) to a vertical position with the cord remaining straight?"
Homework Equations
##v^2=v_0^2+2as##
For part (a) ##s=L(1-cos\theta)##
For part (b) ##s=L(cos\theta)##
##L=\frac{5}{4}m##
##\theta_0=40°##
##v_0=8\frac{m}{s}##
The Attempt at a Solution
(a)
##v^2=v_0^2+2as=v_0^2+2gL(1-cos\theta)##
##v=\sqrt{v_0^2+2gL(1-cos\theta)}=\sqrt{(8\frac{m}{s})^2+2(\frac{98}{10}\frac{m}{s^2})(\frac{5}{4}m)(1-cos(40°))} = \sqrt{64+\frac{49}{2}(1-cos(40°))}\frac{m}{s}=8.35\frac{m}{s}##
(b)
##v^2=v_0^2+2as##
##0=v_0^2-2gL(cos\theta)##
##v_0=\sqrt{2gL(cos\theta)}=\sqrt{(\frac{49}{2})(cos(40°))}\frac{m}{s}=4.33\frac{m}{s}##And here is what I had trouble with:
I learned in Calc. III that for circular motion to happen, the vector for acceleration must be equal and normal to the vector for velocity. But gravity is always pointing down; it is normal to the velocity vector only at the top and the bottom of the circle. What's more, velocity must be constant throughout the rotation. I started with calculating the work that gravity does in half a rotation.
##W_g=\int_{-L}^{L}(mg)dy=2(mg)L##
Then I attempted to do the same for the circular acceleration:
##W_c=\int_{0}^{L}\frac{mv^2}{r}dy=\int_{-\frac{π}{2}}^{\frac{π}{2}}\int_{0}^{L}(csc\theta)mv^2drd\theta=\int_{-\frac{π}{2}}^{\frac{π}{2}}(Lmv^2)(csc\theta)d\theta=2(Lmv^2)##
I don't get the correct numbers when I equate these two energy values:
##2(mg)L=2(mv^2)L##
##v≠\sqrt{g}##
Can anyone help me? Thanks.