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Physics - Tension and angular velocity
Expert: James J. Kovalcin - 11/5/2009
Question
A conical pendulum is formed by attaching a 420 g ball to a 1.0 m long string, then allowing the mass to move in a horizontal circle of radius 22 cm. The figure below shows that the string traces out the surface of a cone, hence the name.
a) What is the tension in the string?
b) What is the ball's angular velocity, in rpm?
Answer
Because the mass of this pendulum is moving in a circular path it is accelerating and therefore its motion is described by Newton's 2nd Law.
Therefore, the sum of the external forces SF acting on this mass must be equal to the mass of the object m multiplied by the resulting acceleration a: SF=m*a
Since the object is moving in a circular path the acceleration of object must be centripetal a=v^2/R and therefore Newton's 2nd Law becomes: SF=m*[v^2/R]
There are two forces acting on this mass: the force of gravity Fg downward and the Tension in the string Ft acting upward and diagonally towards the support point for the pendulum. Since according to Newton's 2nd Law the sum of the forces IN THE DIRECTION of the acceleration is equal to the product of mass and acceleration and since the centripetal acceleration is always directed towards the center of the circular path in which the object is moving you will need to find what forces or parts of the external forces are exerted in the direction of the acceleration. For this conical pendulum the center of the circular path along which the pendulum is moving is the center of a horizontal circle. Therefore, the centripetal acceleration is also horizontal. The force of gravity is directed straight downward and, therefore, contributes nothing to the centripetal acceleration. On the other hand the tension in the string does have a horizontal component. This horizontal component will be equal to the tension in the string Ft multiplied by the sine of the angle between the vertical and the string of the pendulum. From the information given in the problem this angle can be found using the sine function where the opposite side is equal to the radius of the circular path and the length of the pendulum is the length of the pendulum. Therefore: sinQ=opp/hyp=22cm/100cm and the angle Q=sin^(-1)*[22/100]=12.7deg.
Now that you know the angle of the pendulum you can next determine the tension in the string supporting the tension by using the weight of the object that makes up the pendulum and the angle Q using the cosine function:
cos(Q)=adj/hyp=m*g/Ft solve for Ft=m*g/cos(Q)=0.42*9.8/cos(12.7)=4.22N
With this you can find the horizontal component of the tension:
Fh=Ft*sin[12.7]=4.22*sin(12.7)=0.93N
This horizontal component is the only force causing the pendulum to move along its circular path and can therefore be inserted into Newton's 2nd Law:
SF=Ft=0.93=m*[v^2/R]=0.42*[v^2/0.22] solving for v=sqrt[0.93*0.22/0.42]=0.70m/s
Finally you can determine the angular velocity w of the pendulum by dividing the linear velocity v of the pendulum by the radius R of the corresponding circular path:
w=v/R=0.70m/s/0.22m=3.2rad/s
To get this in RPM use dimensional analysis:
w=3.2rad/s*[1rev/6.28rad]*[60s/1min]=30.6RPM
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