The platform is rotating about the vertical axis such that at any instant its angular position is u = (4t3/2) rad, where t is in seconds. A ball rolls outward along the radial groove so that its position is r = (0.1t3) m, where t is in seconds. Determine the magnitudes of the velocity and acceleration of the ball when t = 1.5 s.

Explanation:

angular position u = 4

radial position r = .1t³

s = length of arc = ru

= 4 x .1t³

= .4

ds/dt = .4 x 9/2 x

tangential velocity

Vt = 1.8 x  

= 7.44 m /s

tangential acceleration  At =1.8x 7/2 x

= 17.36 m /s²

radial velocity Vr = dr/dt

= .3t²

= .3 x 1.5 ²

= .675 m/s

radial acceleration Ar = dVr / dt

= .6t = .6 x 1.5

= .9 m /s²

total velocity = √( Vt² + .Vr²) = √( 7.44² + .675²) =  7.47 m /s

total acceleration = √ (At² + Ar ²)  =      √ 17.36² + .9² = 17.38 m /s²

small = small current large mag large current

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west