Problem 10.81 Part A A small sphere of radius ro = 1.2 cm rolls without slipping

Question

Problem 10.81 Part A A small sphere of radius ro = 1.2 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 Express your answer using two significant figures. as shown, what will be its speed. u= m/s Submit My Answers Give Up 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. Figure 1 011 135 Submit My Answers Give Up Continue

Explanation / Answer

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

= 135° = 2.356194 radian.

The altitude of the sphere from the launch point = 24.3*cos45 = 17.18 cm

Potential energy of the sphere = mg*(0. 0.1718)

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

mg (0. 1718) = 0.5mv^2+ 0.5I (v^2/r^2).

I for sphere = 2/5mr^2

9.8* 0. 1718 = 0.5v^2+ 0.2v^2 = 0.7v^2

v = 1.5508 m/s at an angle 45° to the horizontal.

From the launch point the vertical displacement to the ground is

- (24.3 - 17.18) cm = - 7.12 cm = - 0.0712 m

Vertical upward velocity is 1.5508*sin 45 = 1.096 m/s.

Acceleration = -9.8 m/s^2

Time to traverse this displacement is found using

-0.0712 = 1.096t - 4.9t^2

t = 0.276 s

In this time it moves horizontally a distance of 1.508*cos 45*0.276 = 0.3030m

Adding the initial horizontal distance from the bottom to the launch point 0.243*sin 45,

we get 0.1718 + 0.3030 = 0.4748 m

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