An unbalanced baton consists of a 50-cm-long uniform rod of mass 210 g. At one e

Question

An unbalanced baton consists of a 50-cm-long uniform rod of mass 210 g. At one end there is a 10-cm-diameter uniform solid sphere of mass 570 g, and at the other end there is a 8.0-cm-diameter uniform solid sphere of mass 830 g. (The center-to-center distance between the spheres is 59 cm.) Where, relative to the center of the light sphere, is the center of mass of this baton? cm 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? magnitude m/s direction downward What is the net external force on the baton while in the air? N What is the baton's acceleration 1.5 s following its release? magnitude m/s^2 direction downward Use the definition of center of mass to calculate it. Apply constant-acceleration kinematics to the baton to predict its motion.

Explanation / Answer

Xcm = (m1x1cm + m2x2cm + m3x3cm)/(m1+m2+m3)

Xcm = (570 * 0 + 210*30 + 830 * 59) / 570 + 210 + 830

Xcm = 34.33 cm

part b )

v = u - gt

v = 10 - 9.8 * 1.5 = -4.7 m/s

downwards

part c )

F = mg

1g = 10^-3 kg

Fnet = (m1+m2+m3)*9.8 = 15.778 N

part d )

9.8 m/s^2 ( equal to g becuase baton acts like a point )

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