2. ## Re: Mechanics (Pendulum) Question

b)
$cot\alpha = \frac{g}{w^2r}$ (dividing the 2 equations from part a)

$=\frac{(R^2-r^2)^{0.5}}{r}=\frac{g}{w^2r}$ (from triangle formed by pendulum)

remove r at bottom and divide by R to get answer

c)
cos a is between a certain region; where it is equal or less than 1 or greater or equal to min value of cosa which occurs when a is max;
$\frac{R-h}{R} \leqslant \frac{g}{w^2R}\leq 1$

$w^2(R-h) \leqslant g\leq Rw^2$

3. ## Re: Mechanics (Pendulum) Question

Originally Posted by mrbunton
b)
$cot\alpha = \frac{g}{w^2r}$ (dividing the 2 equations from part a)

$=\frac{(R^2-r^2)^{0.5}}{r}=\frac{g}{w^2r}$ (from triangle formed by pendulum)

remove r at bottom and divide by R to get answer

c)
cos a is between a certain region; where it is equal or less than 1 or greater or equal to min value of cosa which occurs when a is max;
$\frac{R-h}{R} \leqslant \frac{g}{w^2R}\leq 1$

$w^2(R-h) \leqslant g\leq Rw^2$
why is $\frac{R-h}{R}$ the lower bound? thanks

4. ## Re: Mechanics (Pendulum) Question

when a is max; cosa is minimum. a is max when it is right on the edge of the bowl; so height of ball is h; we take R-h to find the adjacent side of the triangle. hence R-h/R is equal to cosa at its minimum value

5. ## Re: Mechanics (Pendulum) Question

Originally Posted by mrbunton
when a is max; cosa is minimum. a is max when it is right on the edge of the bowl; so height of ball is h; we take R-h to find the adjacent side of the triangle. hence R-h/R is equal to cosa at its minimum value
rip, i completely missed that the vertical length is R as well. thanks again!

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