Answer:
(a) The acceleration is 0.608 [tex]\frac{m}{s^{2} }[/tex]
(b) The minimum value of coefficient of static friction is 0.062
Explanation:
Given:
Angular frequency [tex]\omega = 33 \times \frac{2\pi }{60} = 3.454[/tex] [tex]\frac{rad}{s}[/tex]
Distance between axis of rotation and watermelon [tex]r = 5.1 \times 10^{-2}[/tex] m
(a)
For calculating acceleration,
From the formula of centripetal acceleration,
[tex]a_{c} = r \omega ^{2}[/tex]
[tex]a_{c} = 5.1 \times 10^{-2} \times (3.454) ^{2}[/tex]
[tex]a_{c} = 0.608[/tex] [tex]\frac{m}{s^{2} }[/tex]
(b)
For finding the minimum value of the coefficient of static friction,
[tex]F = ma_{c}[/tex]
But here only friction force act
[tex]\mu_{s} mg = ma_{c}[/tex]
Where [tex]g = 9.8 \frac{m}{s^{2} }[/tex]
[tex]\mu _{s }= \frac{a_{c} }{g}[/tex]
[tex]\mu _{s} = \frac{0.608}{9.8}[/tex]
[tex]\mu _{s} = 0.062[/tex]
Therefore, the acceleration is 0.608 [tex]\frac{m}{s^{2} }[/tex] and minimum value of coefficient of static friction is 0.062
