Linear Motion Ball
Tuesday, September 13th, 2011
Linear Motion Ball 
physics problem dealing with both linear and rotational motion…please help!?
A marble rolls on the track shown in the figure below, with hB = 20 cm and hC = 16 cm. If the marble has a speed of 3.0 m/s at point A, what is its speed at points B and C?
there’s a picture:
point hA has height of 0 cm. hB is the peak. and velocity does not stop at hC
moment of inertia of ball is 2/5mr^2
done this problem twice and got 3.35 for vB but it’s wrong…
please help, show work and explain! thank you.
ΔE = 0
½ m VA² + ½ (2/5) m r² (VA/r)² – ½ m VB² – ½ I ωB² – m g hB = 0
½ (VA)² + ½ (2/5) (VA)² – ½ VB² – ½ (2/5) (VB)² – g hB = 0
½ (3)² + ½ (2/5) (3)² – ½ VB² – ½ (2/5) (VB)² – (9.8)(0.2) = 0
VB = 2.489979919597746475 m/s
and at C point,
½ (VA)² + ½ (2/5) (VA)² – ½ VC² – ½ (2/5) (VC)² – g hC = 0
½ (3)² + ½ (2/5) (3)² – ½ VC² – ½ (2/5) (VC)² – (9.8)(0.16) = 0
VC = 2.6 m/s