For Physics lovers!?!?

Question

A 0.45-kg ball is attached to the end of a cord of length 1.4 m. The ball is whirled in a circular path in a horizontal plane. The cord can withstand a maximum tension of 57.0 N before it breaks. What is the maximum speed the ball can have without the cord breaking?

Can someone please explain to me how to do this? I would be very thankful. Please help me. Thank you.

Answer 1

centrpetal force =mV^2/r
57.0=0.45*(V)^2/(1.4)

v=13.3ms^-1.

Answer 2

Let the cord make an angle θ above the line joining the centre of rotation and the ball at a time when the tension in it is maximum = 57 N.
Vertical component of tension in the cord T = T sin θ balances the weight, mg = 0.45 * 9.8 = 4.41 N
=> 57 sin θ = 4.41
=> θ = 4.4373°

Horizontal component of tension T = T cos θ provides the necessary centripetal force = mv^2 / r cos θ
=> T cos θ = mv^2 / r cos θ
=> v = √ [(Trcos^2 θ) / m]
= cos (4.4373°) √ [ (57 * 1.4) / (0.45) ]
= 13.3 m/s.

[ Note: In this problem, the value of T is given to be so high that angle θ is very small. In such a case, an approximate method taking θ = 0° ( i.e., cos θ = 1) will give almost the same answer. But if some examiner gives such a low value of T that we cannot take cos θ = 1, then the above rigorous method only will give the correct answer. ]

Answer 3

To move in a circle, the rock must experience a centripetal force equal to

m v^2 / r

The string provides this centripetal force. The largest tension that the string can handle is 57.0 N, so

57.0 N = m v^2 / r

You know m and r; solve for v.

Answer 4

Physics is a great subject and you a correct you can learn it if you dont enjoy studing it. Okay i'l suggest you hang with friends who enjoys it as well when you read something share with them this way u'll want to learen a new thing everyday. Last year i was doing my class 12 we created a study group we were five we treated the papers and argued alot bellieve me you it helps. We all passed with 80%+ i myself managed to obtain 95%. Now i'm doing my 1st year in electrical engineering and i LOVE it. Try to apply physics in real life problem and discuss alot

Answer 5

the motion can be seen as a conical pendulum. let angle be θ of cord with the vertical line so that it acts as semi-vertical angle of cone formed at hand.

2 forces act, mg downwards, T (tension) along cord at hand
tking components of T
T cos (θ) = mg
T sin (θ) = provides centripetal force = mv^2/r
sin θ = r/L or r = L sin θ
===========
given T =57 N
cos (θ) = 0.45*9.8/57 = 0.07737 >>>> (θ) = 85.5 degree
sin (θ) = 0.997
v^2 = r *T sin (θ)/m = (L T/m) sin ^2 θ
v^2 = (1.4*57/0.45)(0.997)^2
v = 13.28 m/s