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oa The Expert TA | Human-like Gr x C secure https://usfi6in.theexpertta.co TakeTu Homework 4 Begin Date: 9/19/2018 12:01:00 AM --Due Date: 9/26/2018 11:59:00 PM End Date: 9/26/2018 11:59:00 PM (10%) Problem 5: On the International Space Station an object with mass m-350 g is attached to a massless string L = 0.78 m. The string can handle a tension of T-6.1 N before breaking. The object undergoes uniform circular motion, being spun around by the string horizontally Assignment Status Click here for detailed view Problem Status 1 Completed 2 Completed 3 Completed 4Completed 5 Partial * what is the maximum speed v the mass can have before the string breaks? Give your answer in units of m/s. Grade Summary Deductions 0% 100% sinO cotanasin) Submissions Attempts remaining: 6 (0% per attempt) detailed view cosO tan() acosO atan) acotan sinhO coshO tanh0 cotanh) ODegrees Radians 0% Submit Hint I give up! Hints; 3 for a 0% deduction. Hints remaining: 1 Feedback: 0% deduction per feedback. Start with a free body diagram for the rock. What force is providing the centripetal force? -Can you solve for the speed by oquating the tension to the centripetal force?Explanation / Answer
Hello,
The body moving in a circular motion has to have a centripetal force acting on it which is responsible for its curved motion. Here in this case the tension of the string provides the necessary centripetal force for the object to spun around it.
Centripetal force F = mv2/r ; where m- mass of the object
v- speed of the object in circular motion
r- radius of the circular path
As centripetal force is dependant on the speed v, this means that tension in the string adjusts just as enough as is demanded by the speed attribute in F. Hence at the point of maximum tension in the string, velocity of the body will be maximum possible that the string can afford without breaking itself.
Therefore, Fmax=Tmax
==> m(vmax)2/r = Tmax = 6.1N
==> 0.35*v2/0.78 = 6.1
==> v2= 6.1*0.78/0.35
==> vmax = 3.68m/s
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