Physics - Circular Motion Word Problem Help Needed?

Question

This problem is 5 parts, here is the background information:

A flea is at point "A" on a horizontal turntable that is 14.7 cm from the center. The Turntable is rotating at 32.5 rev / min in the counterclockwise direction. The flea jumps vertically upward to a height of 7.29 cm and lands on the turntable at point "B" . Place the coordiante origin at the center of the turntable with the positive x-axis fixed in space and initially passing through "A". The acceleration of gravity is 9.8 m/s^2.

A.) Find the linear displacement of the flea ( Answer in meters)

B.) Find the angle through which point A has rotated when the flea lands (Answer in degrees)

C.) Find the distance of point "B" from the origin when the flea lands (Answer in meters)

D.) What angle does the radius drawn through "B" make with the x-axis when the flea lands (Answer in degrees)

E.) What is the linear speed of the flea after it lands? (Answer in m/s)

Answer 1

Ok, here's how you do it:

First you get the datta

x = 14.7 cm = 0.14m

y = 0 m

z = 0 m

the frecuency is 32.5 rev/min

the height it gets:

z = 7.29 cm = 0.0729m

and the acceleration is 9.8 m/s**2

A)

Now you transform the frecuency to angular velocity

w = (32.5 rev/min)*(2*pi rad/1rev)*(1min/60s)=3.4 rad/s

that way you can get the velocity in y

Vy = wr = 3.4 rad/s * 0.147m = 0.5 m/s

now you use the next equation solved for t

t = sqrt (2z/g) = sqrt(2*0.0729m/9.8m/s**2) = 0.122s

since that time is what it takes the flea to get to the top, then you multiply it by two to get it's total jump time

0.122s*2 = 0.244s

and then you can get the linear displacement of the flea. Since the velocity in y is going to be constant, then:

y = vt = 0.5 m/s * 0.244s = 0.122 m

B)

You multiply the angular velocity per the time it takes the fly to jump to find the angle that has rotated point A

Wt = 3.4rad/s*0.244s = 0.83 rad

since it's in rad, we pass it to degrees

0.83 rad * (360º/2pi rad) = 47.6º

C) the x coordinate is going to remain the same 14.7 cm and the y coordinate is the one you cot on letter A, so using pitagore.

l = sqrt((0.147m)**2+(0.122m)**2) = 0.1910 m

D) in order to get the angle you use the tan-1 function

tan-1(0.122/0.147)=39.7º

E) again, using the angular velocity and the displacement of the flea from the center you get the linear velocity of the fly

V = Wr = 3.4 rad/s*0.191 m = 0.649 m/s

And that's it

When i use ** it means squared and when i use sqrt is squared root.

hope this helps.

Answer 2

a) first find the initial vertical speed of the jump using the forumula: vy_final^2=vy_initial^2-2gd (d = 0.0729m and v_final=0m/s)...v_initial=1.19m/s.
now, plug this into the formula y=(v_initial)t-0.5gt^2 where y=0, to get the time the flea spends in the air...t=0.244s.
now, find the linear speed. the flea will travel at constant linear velocity at a tangent from the turntable at the instant he jumps. the velocity of the turntable at the radius of 14.7cm is found by using the formula: v_linear=circumference/period or circumference*frequency; v_linear=2*radius*pi*(32.5rev / 60sec)...v_linear=0.5m/s.
if the flea travels at constant linear velocity while in the air for 0.244s, his displacement is 0.5m/s * 0.244s = 0.122m.
b) the angle the turntable rotates in 0.244s is found using the expression: 360deg/rev*(32.5rev/60sec) * 0.244s = 47.6deg
c)using pythagorus, where 14.7cm is the adjacent side, 0.122m is the opposite side, the hypotenuse is 0.191m, which is the distance from B to origin
d)using trig in one of several ways, tan(angle) = 0.122/0.147. angle is 39.7deg
e)the linear speed of the flee is the same as it was when it first jumped, since it travels at constant velocity in the linear/horizontal tangential direction, 0.5m/s.

Answer 3

the concept of the car sliding and overturning vary sliding can clearly be done with the concept of friction but the overturning needs to be solved by the concept of torque and needs additional data. the forces with respect to car are its own weight that contributes to frictional force and the centripetal force you have got to consider these both with respect to sliding of the car because the centripetal reaction often confused with the centrifugal force tries to slide the car where as the frictional force tries to prevent that since the car should not slide the maximum centripetal reaction that is directed outwards of the circular track should not exceed the frictional force here remember that centripetal force is acting towards the center but its reaction component is acting outwards ..okay and you always take the reaction component now the frictional force is opposite to the direction of the applied force that is inwards coming to the equations the frictional force is given = 0.8M the centripetal reaction equals the centripetal force that is mv^2/r r refers to the radius of the track.M the mass of the car and v is the velocity dont confuse withe the formula its v square not 2 times okay r=20m 0.8M=Mv^2/20 m gets canceled 0.8=v^2\20 0.8*20=v^2 16=v^2 so v=4m/s^2 your next question needs to be explained by a diagram which is not possible to draw it on yahoo answers anyways the concept is the same the bob of the pendulum swings outwards making some angle with the vertical due to centripetal reaction the forces acting are the centripetal force and the weight but here both are acting at some angles hence consider it to be x now resolve that into components wrt angle made with the vertical you will have tan(x) draw it on the paper and it will become easy for you good luck