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Dvinal [7]
2 years ago
8

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.
Physics
1 answer:
Svetach [21]2 years ago
3 0

Answer:

Explanation:

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

radial position r = .1t³

s = length of arc = ru

= 4t^{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

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irinina [24]
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1 year ago
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