A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 1.7 cm from the axis of rotation. (a) Calculate the acceleration of the seed, assuming that it does not slip. Rest below but can anyone please help me?
(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 0.29 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.
I don't understand how to fully go about solve this problem. I've attempted it f=and for the most part gotten all parts wrong.
(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 0.29 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.
I don't understand how to fully go about solve this problem. I've attempted it f=and for the most part gotten all parts wrong.
1 Answer
All work in SI units.
Explanation:
(a) The angular velocity in is
#omega=2pif#
#=>omega=2pixx33/60=1.1pi\ radcdot s^-1#
Consequently, the centripetal (radial) acceleration is
#a_r=Romega^2#
#=>a_r=1.7/100xx(1.1pi)^2#
#=>a_r=0.20\ ms^-2#
(b) For the seed of mass
#ma_r<=f_s#
Also#f_s=f_(s(max))=mu_smg#
where#mu_s# is coefficient of static friction and#g# is acceleration due to gravity.
Therefore we get
#ma_r=mu_smg#
#=>mu_s=a_r/g#
Inserting values we get
#mu_(smin)=0.20/9.81#
#=>mu_(smin)=0.0207#
(c) We know that tangential acceleration
#a_t = Ralpha#
where#alpha# is angular acceleration.
also
Angular acceleration can be found from
#alpha=(Deltaomega)/(Deltat)#
Inserting given values we get
#alpha=(1.1pi-0)/(0.29)=11.916\ rad cdot s^-2#
This gives us
#a_t = 1.7/100xx11.916=0.203\ ms^-2#
Now total acceleration
#a_T=sqrt(a_t^2+a_r^2)#
#=>a_T=sqrt((0.203)^2+(0.20)^2)=0.285\ ms^-2#
Inserting this value in the already derived expression above we get
#mu_(smin)=0.285/9.81#
#=>mu_(smin)=0.029#