Answer:
a) [tex]a_{r} = 0.275\,\frac{m}{s^{2}}[/tex], b) [tex]\mu_{s} = 0.028[/tex], c) [tex]\mu_{s} = 0.036[/tex]
Explanation:
a) The linear acceleration of the watermelon seed is:
[tex]a_{r} = \omega^{2}\cdot r[/tex]
[tex]a_{r} = \left[\left(33\,\frac{rev}{min} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\right]^{2}\cdot (0.023\,m)[/tex]
[tex]a_{r} = 0.275\,\frac{m}{s^{2}}[/tex]
b) The watermelon seed is experimenting a centrifugal acceleration. The coefficient of static friction between the seed and the turntable is calculated by the Newton's Laws:
[tex]\Sigma F = \mu_{s}\cdot m\cdot g = m\cdot a[/tex]
[tex]a = \mu_{s}\cdot g[/tex]
[tex]\mu_{s} = \frac{a}{g}[/tex]
[tex]\mu_{s} = \frac{0.275\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]\mu_{s} = 0.028[/tex]
c) Angular acceleration experimented by the turntable is:
[tex]\alpha = \frac{\omega-\omega_{o}}{\Delta t}[/tex]
[tex]\alpha = \frac{3.456\,\frac{rad}{s}-0\,\frac{rad}{s} }{0.36\,s}[/tex]
[tex]\alpha = 9.6\,\frac{rad}{s^{2}}[/tex]
The tangential acceleration experimented by the watermelon seed is:
[tex]a_{t} = \left(9.6\,\frac{rad}{s^{2}} \right)\cdot (0.023\,m)[/tex]
[tex]a_{t} = 0.221\,\frac{m}{s^{2}}[/tex]
The linear acceleration experimented by the watermelon seed is:
[tex]a = \sqrt{a_{t}^{2}+a_{r}^{2}}[/tex]
[tex]a = \sqrt{\left(0.221\,\frac{m}{s^{2}} \right)^{2}+\left(0.275\,\frac{m}{s^{2}} \right)^{2}}[/tex]
[tex]a = 0.353\,\frac{m}{s^{2}}[/tex]
The minimum coefficient of static friction is:
[tex]\mu_{s} = \frac{0.353\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]\mu_{s} = 0.036[/tex]
