A 90 g bead on a 60 cm long string is swung counter clockwise in avertical circl

Question

A 90 g bead on a 60 cm long string is swung counter clockwise in avertical circle about a point 200 cm above the floor. The tensionin the string when the bead is at the very bottom of the circle is3.4 N. A very sharp knife is suddenly inserted, as the stringreaches 180 degrees perpendicular to the floor to cut the stringdirectly below the point of support. In other words it's cut justas it's pointing straight down and swing to the right. How far tothe right of the center of the circle does the ball hit thefloor?

Thanks for your help!

Explanation / Answer

T = m g + m v2 / R   the tension in thestring at the bottom of the circle v = (R ( T - mg) / m) h = 1/2 gt2          t = (2 h / g)    time forparticle to fall to floor h = (H - R) = .2 - .06 = .14 m    distanceparticle falls Horizontal distance particle travels s = v t = (R( T - mg) / m) * (2 * .14 / g) Substitute and solve

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