A turntable is rotating at a 33.33 rev/min. A watermelon seed is on the turntabl

Question

A turntable is rotating at a 33.33 rev/min. A watermelon
seed is on the turntable 6.0 cm from the axis of rotation. (a)
Calculate the acceleration of the seed, assuming that it does
not slip. (b) What is the minimum value of the coefficient
of static friction between the seed and the turntable if the
seed is not to slip? (c) Suppose that the turntable achieves
its angular speed by starting from rest and undergoing a
constant angular acceleration for 0.25 s. Calculate the
minimum coefficient of static friction required for the seed
not to slip during the acceleration period.

Explanation / Answer

I assume a steady rotation. Convert speed from RPM into an omega in rad/sec. (A) The acceleration of the watermellon seed is centripetal, since it is traveling in a circular path. a_cent = omega^2 * R (B) Friction holds it in place. Friction is proportional to normal force, via roughness coefficient mu. Normal force is equal and opposite weight in order that it doesn't fall through turntable surface. F = mu*m*g mu*m*g = m*a_cent mu*m*g = m*omega^2 * R Solve for mu: mu = omega^2*R/g (C) The watermelon seed also undergoes tangential acceleration, a perpendicular component to centripetal acceleration. Tangential acceleration: a_tan = alpha*R alpha = omega/?t thus, a_tan = omega*R/?t Combine accelerations in Pythagorean theorem combination: a_net = sqrt(a_tan + a_cent) a_net = sqrt(omega*R/?t + omega^2*R) In order that the seed remains in place without slipping, this acceleration must be caused by friction. mu*m*g = m*a_net mu*g = sqrt(omega*R/?t + omega^2*R) Result for mu: mu = sqrt(omega*R/?t + omega^2*R)/g --- Results for question 3: Data: omega := 10*Pi/9 rad/sec; R:=0.054 m; ?t:=0.55 sec; g:=9.8 N/kg; Part A) a_cent = omega^2 * R a_cent = 0.6556 m/sec^2 Part B) mu = omega^2*R/g mu = 0.0669 Part C) mu = sqrt(omega*R/?t + omega^2*R)/g mu = 0.314

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.