HW8-phy2048-Spring-2014-solution

# 10 m 3 b now this becomes a projectile motion of the

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Unformatted text preview: 2 Iω 2 . Here we use v to denote the speed of its center of mass and ω is its angular speed — at the moment it leaves the roof. Since (up to that moment) the ball rolls without 1 sliding we can set v = Rω = v where R = 0.10 m. Using I = 2 MR 2 (Table 10- 2(c)), conservation of energy leads to 1 1 1 1 3 Mgh = Mv 2 + I ω 2 = MR 2ω 2 + MR 2ω 2 = MR 2ω 2 . 2 2 2 4 4 The mass M cancels from the equation, and we obtain 14 1 4 ω= gh = ( 9.8 m s2 )( 3.0 m ) = 63 rad s . R3 0.10 m 3 (b) Now this becomes a projectile motion of the type examined in Chapter 4. We put the origin at the position of the center of mass when the ball leaves the track (the “initial” position for this part of the problem) and take +x leftward and +y downward. The result of pa...
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