Question

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.

Answer 1

Answer:

Explanation:

angular position u = 4$t^{3/2}$

radial position r = .1t³

s = length of arc = ru

= 4$t^{3/2}$ x .1t³

= .4 $t^{9/2}$

ds/dt = .4 x 9/2 x $t^{7/2 }$

tangential velocity

Vt = 1.8 x  $1.5^{7/2 }$

= 7.44 m /s

tangential acceleration  At =1.8x 7/2 x $1.5^{5/2}$

= 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