Problem 10.81 Part A A small sphere of radius ro = 1.5 cm rolls without slipping

Question

Problem 10.81 Part A A small sphere of radius ro = 1.5 cm rolls without slipping on the track shown in the figure whose radius is Ro-25.5 cm. The sphere starts rolling at a height Ro above the bottom of the track. (Figure 1) When it leaves the track after passing through an angle of 135°as shown, what will be its speed. Express your answer using two significant figures. v= 1.5 m/s Submit My Answers Give Up Correct Part B When it leaves the track after passing through an angle of 135° as shown, at what distance D from the base of the track will the sphere hit the ground? Express your answer using two significant figures Submit My Answers Give Up Figure 1of 1 Continue Ro 135

Explanation / Answer

A)

The center of mass moves through arc of length = (Ro - r)*

=135°= 2.356194 radian.

The altitude of the sphere from the launch point = (25.5-1.5) cos 45 =16.97 cm

Potential energy of the sphere = mg (0. 1697)

This is equal to the translational k.e and rotational k.e of the sphere at the launch point.

mg (0. 1697) = 0.5mv^2+ 0.5I (v^2/r^2). I for sphere = 2/5mr^2

9.81* 0. 1697 = 0.5v^2+ 0.2v^2 = 0.7v^2

v = 1.5 m/s at an angle 45° to the horizontal.

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B)

From the launch point the vertical displacement to the ground is

- ((25.5 – 1.5) -16.97) cm = -0.0703 m

Vertical upward velocity is 1.542 sin 45 = 1.09 m/s.

Acceleration = -9.81 m/s^2

Time to traverse this displacement is found using

-0.0703 = 1.09t - 4.905t^2

t = 0.274 s

In this time it moves horizontally a distance of (1.542 cos 45)*0.274 = 0.2988m

Adding the initial horizontal distance from the bottom to the launch point i.e.

(0.255 – 0.015) * sin 45 = 0.1697

, we get

0.1697 + 0.2988 = 0.47 m

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