An unbalanced baton consists of a 50-cm-long uniform rod of mass 190 g. At one e

Question

An unbalanced baton consists of a 50-cm-long uniform rod of mass 190 g. At one end there is a 10-cm-diameter uniform solid sphere of mass 470 g, and at the other end there is a 8.0-cm-diameter uniform solid sphere of mass 810 g. (The center-to-center distance between the spheres is 59 cm.)

(a) Where, relative to the center of the light sphere, is the center of mass of this baton?
  cm

(b) If this baton is tossed straight up (but spinning) so that its initial center of mass speed is 10.0 m/s, what is the velocity of the center of mass 1.5 s later?

(c) What is the net external force on the baton while in the air?
  N

(d) What is the baton's acceleration 1.5 s following its release?

Explanation / Answer

a)

CM distance from light sphere

CM = 810 x 59 + (30x190) / 470+810+190

CM = 36.387 cm

b)

velocity

V = u - gt

V = 10- g (1.5

V = -4.7 m/s ( negative indicates downward )

c)

acceleration

a = g = 9.8 m/sec^2

d)

Net force

F = total mass x a

F = 14.406 N

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