A 1.51 kg disc is attached to the end of a stringw...

Answers ( 5 )

  1. shaina pijnappel
    shaina pijnappel
    KEVIN42 DECIDES TO PLAGERIZEKHM
  2. herbert hall
    herbert hall
    In thiscase, the tension in the string provides the centripetal force tokeep the disc moving in a circle. If you spin too fast, the forceneeded to keep the disk moving in a circle exceed the strength ofthe string. So, we use the equation for centripetal force andrearrange to find the tangential speed the will exceed the tensilestrngth of the string.
    Fc=T=mv^2/r 143.7 N<1.51 kg*v^2/.340m        v <?(143.7N*.340 M/1.51 kg)=5.69 m/s
  3. lorenzo roche
    lorenzo roche
    Now the object is moving in a vertical circle so gravity is afactor. Both gravity and the string are pulling on the disk.Gravity will always pull with a force of m*g or 1.51kg*9.8m/s^2=15.8 N Next, you need to find the total force needed tokeep the disk moving in a circle with can be found by usingm*v^2/r. The force needed to keep the disk moving in a circle- the force of gravity acting on the disk, tells you how much forcethe string needs to be. Think of it as the string plus gravity pullthe disk in a circle when it goes over the top. When is goes pastthe bottom of the circle, the string must pull it into the circleplus the disk's weight. More tension is needed at the bottombecause gravity isn't helping in this case.
  4. julia roger
    julia roger
    The second part is "
  5. enni salonen
    enni salonen
    In this case, the tension in the string provides thecentripetal force to keep the disc moving in a circle. If you spintoo fast, the force needed to keep the disk moving in a circleexceed the strength of the string. So, we use the equation forcentripetal force and rearrange to find the tangential speed thewill exceed the tensile strngth of the string.
    Fc=T=mv^2/r 143.7 N<1.51 kg*v^2/.340m        v <?(143.7N*.340 M/1.51 kg)=5.69 m/s

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