1)How fast is the mass moving at the bottom of its path? 2)What is the magnitude

Question

1)How fast is the mass moving at the bottom of its path?

2)What is the magnitude of the tension in the string at the bottom of the path?

3)If the maximum tension the string can take without breaking is Tmax = 680 N, what is the maximum mass that can be used? (Assuming that the mass is still released from the horizontal and swings down to its lowest point.)

4)Now a peg is placed 4/5 of the way down the pendulum’s path so that when the mass falls to its vertical position it hits and wraps around the peg. As it wraps around the peg and attains its maximum height it ends a distance of 3/5 L below its starting point (or 2/5 L from its lowest point).
How fast is the mass moving at the top of its new path (directly above the peg)?m/s



5)Using the original mass of m = 8 kg, what is the magnitude of the tension in the string at the top of the new path (directly above the peg)?

Explanation / Answer

Given Mass m = 8 kg Length of the rope l = 2.17 m a) From law of conservation of energy              ( 1 / 2 ) m v 2 = m g l Speed of mass at the bottom of path                              v = ( 2 g l ) 1/2                                    = ( 2 *9.8 m/s2 * 2.17 m ) 1/2                                  = 6.521 m/s __________________________________________ b) Tension in the string at the bottom of path is                           T = m v 2 / l + mg                               = m ( v 2 / l + g )                               = ( 8 kg ) ( 6.5212 / 2.17 m + 9.8 m/s2 )                                = 235.2 N ________________________________________________ c) Maximum tension in the string T = 680 N      Maximum mass                       m = T / ( ( v 2 / l + g )                            = ( 680 N ) / [ (6.521 m/s )2 / 2.17 m    + 9.8 m/s2 ]                             = 23.13 kg _______________________________________________________                               = ( 8 kg ) ( 6.5212 / 2.17 m + 9.8 m/s2 )                                = 235.2 N ________________________________________________ c) Maximum tension in the string T = 680 N      Maximum mass                       m = T / ( ( v 2 / l + g )                            = ( 680 N ) / [ (6.521 m/s )2 / 2.17 m    + 9.8 m/s2 ]                             = 23.13 kg _______________________________________________________                         

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