What is the velocities, acceleration, position, frequency??? PLEASE SHOW ALL WORK =)?

Question

We've started Circular Motion in physics but I don't understand the concept even after reading the book.please can someone explain how to do this problem?

The lazy susan has been rotating for some time in this problem at a constant angular frequency of 2 sec^-1. At this moment the position of A and B are given as: r(a)=.5 m @ 180 (degrees) and r(b)=.25 m @ 270 (degrees)
(a) At this moment what are the velocities of cans A and B?
can A ___m/s @ ___degrees
can B ___m/s @ ___degrees
(b) At this moment what are the accelerations of cans A and B?
can A ___m/s^2 @ ____degrees
can B ___m/s^2 @ ____degrees
(c) What is the period and frequency for can B?
T= ___sec
f= ___sec^-1
(d) In one second what will be the position, velocity, and acceleration of can B?
r= ___m @ ___degrees
v= ___m/s @ ___degrees
a= ___m/s^2 @ ___degrees
Also, please please show all work with all equations used
Thanks a lot for your help =

Answer 1

Angular frequency must first be converted to angular speed. If the lazy susan rotates two revolutions per second, a point on the rotating surface will traverse the circumference twice every second. The circumference is equal to 2π radians.
Therefore, any point on the surface of the lazy susan has angular speed of 4π rad/second
The translational speed, or tangential speed, is dependent on the radius from the axis of rotation by the equation
ω*r=v
This is intuitive since circumference is related to radius as
2πR
The larger the R, the larger the circumference, and the more distance a particle must traverse every revolution.

to convert radians to degrees, every 2π radians is 360 degrees.
I will assume the lazy susan is rotating counter clockwise.
The
ω=4π rad/s
a) r(a)=0.5 m
θ(a)=180 degrees, or 2π radians

r(b)=0.25 m
θ(a)=270 degrees, or 3π radians
a) The translational speeds are
A: 0.5*4*3.14 m/s
6.28 m/s at 90 degrees below horizontal

B: 0.25*4*3.14
3.14 m/s at 0 degrees to the right

j

Answer 2

A) If x = 310mcos(130t): the fee in front of the 'cos' (i.e. 310metres) is the amplitude, A. this is because of the fact the main important cos could be is a million; so while cos =a million x = max fee (amplitude) = A = 310m. this is no longer 309m! the fee in front of the t is the angular velocity ? . this is in radians in line with 2nd. So ? = 130rad/s sensible formulation to keep in mind are ? =2.pi.f and f = a million/T (or ?=2.pi/T) So f = ?/(2.pi) = 20.7Hz T = a million/f = 0.0483s B) Vmax is given by using -?A (basic to coach by using differentiating x=.310mcos(130t)) overlook approximately minus sign: Vmax = 130x310 = 40,300m/s a = -?^2x for SHM (basic to coach by using differentiating), so amax = -?^2A = a hundred thirty^2 x 310 = 5.24x10^6m/s^2 C) the situation, velocity, and acceleration while t = 0.250 s x=.310mcos(130t)) = 310mcos(130x0.250) = (artwork out for your self with calculator in radian mode) v = dx/dt = -310mx130sin(a hundred thirty x 0.250) (artwork out for your self ...) a = dv/dt = -310mx130^2cos(a hundred thirty x 0.250) ((artwork out for your self ...) in case you have a formulation sheet, verify you will detect and comprehend the appropriate formulation.