contestada

A coin of mass m is placed on top of a horizontal turntable a distance R from the axis of rotation. Gravity is acting vertically downward with magnitude mg and the normal force on the coin from the turntable is acting vertically upward. The coefficient of static friction between the coin and the turntable is µ. If at time t = 0 the turntable is at rest and accelerates with a constant angular acceleration α < µg/R, at what time T will the coin begin to slide on the turntable?

Respuesta :

hafsaabdulhai

Answer:

T= 6.28fR²/µg

Explanation:

a=µg/R

(v-vi)/(t-ti)= µg/R

v: linear velocity of turntable at which coin begins to slip

vi: initial velocity

t: time at which coin begins to slide

ti: time at which turntable started moving

(v-0)/T=µg/R

T=Rv/µg

to find v,

v= wR

here, w is angular velocity and is equal to 2πf where f is frequency of turntable rotation

v= 2πfR

so,

T= R (6.28fR)/µg

T= 6.28fR²/µg

adeoladokun31

The time that the coin begin to slide is 6.28fR²/µg

From the question the formular for time is given below.

T= 6.28fR²/µg

while the angular acceleration is given below.

a=µg/R

And acceleration is (v-vi)/(t-ti)

Therefore,

(v-vi)/(t-ti)= µg/R

V is linear velocity of turntable where the coin begins to slip

Vi is the nitial velocity

T is time at which coin begins to slide

T1 is time at which turntable started moving

(v-0)/T=µg/R

T=Rv/µg

To get V.

v= wR

Where,

w is angular velocity which is equal to 2πf

f is frequency of turntable rotation

v= 2πfR

Therefore,

T= R (6.28fR)/µg

T= 6.28fR²/µg.

What is motion?

Motion refer to process in which an object move or change position.

Motion can be inform of displacement, acceleration, velocity and so on.

Therefore, the time the coin begin to slide is 6.28fR²/µg

Learn more about motion from the link below.

https://brainly.com/question/22921263

Otras preguntas