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In Fig. 6-45, a1.34kgball is connected by means of two massless strings, each of lengthL=1.70m, to a vertical, rotating rod. The strings are tied to the rod with separationd=1.70mand are taut. The tension in the upper string is35N. What are the

(a) tension in the lower string,

(b) magnitude of the net forceFneton the ball, and

(c) speed of the ball?

(d) What is the direction ofF?

Short Answer

Expert verified
  1. Tl is 8.74 N.
  2. Fnet,stis37.9 N.
  3. v is 6.45 m/s.
  4. The direction of Fnet,stris leftward.

Step by step solution

01

Given data

  • Mass of ball,m=1.34 kg.
  • Length of each string,L=1.70 m.
  • The separation between two strings tied to rod,d=1.70 m.
  • Tension in the upper string,Tu=35 N .
02

To understand the concept

Using the concept of centripetal force and applying Newton's second law, we can solve the given problem. Note that the tensions in the strings provide the source of centripetal force.

03

Draw the free body diagram and write the force equations

The given system consists of a ball connected by two strings to a rotating rod. The tensions in the strings provide the source of centripetal force.

The free-body diagram for the ball is shown below. Tuis the tension exerted by the upper string on the ball, Tlis the tension in the lower string, and role="math" localid="1661234901128" mis the mass of the ball. Note that the tension in the upper string is greater than the tension in the lower string. It must balance the downward pull of gravity and the force of the lower string.

We take the+xdirection to be leftward (toward the center of the circular orbit) and+yupward. Since the magnitude of the acceleration isa=v2/R, thexcomponent of Newton's second law is,

Tucosθ+Tlcosθ=mv2R

Wherevis the speed of the ball, andR is the radius of its orbit.
They component is,

TusinθTlsinθmg=0

The second equation gives the tension in the lower string:

Tl=Tumg/sinθ.

04

(a) Calculate the tension in the lower string

Since the triangle is equilateral, the angle isθ=30.0o.

Thus,

Tl=Tumgsinθ

Substitute the values, and we get,

Tl=35.0 N(1.34 kg)(9.80 m/s2)sin30.0Tl=8.74 N

Thus, Tl is 8.74 N.

05

(b) Calculate the magnitude of the net force F¯net on the ball 

The net force in they direction is zero. In thex-direction, the net force has magnitude as:

Fnet,st=(Tu+Tl)cosθ

Substitute the values, and we get,

Fnet,st=(35.0 N+8.74 N)cos30.0oFnet,st=37.9 N

Thus, Fnet,stis37.9 N.

06

(c) Calculate the speed of the ball 

The radius of the path is,

R=Lcosθ

Substitute the values, and we get,

R=(1.70 m)cos30o=1.47 m

Using thisFnet,str=mv2/R, we find the speed of the ball to be,

v=RFnet,strm

Substitute the values, and we get,

v=(1.47m)(37.9N)1.34 kgv=6.45 m/s

Thus, v is 6.45 m/s.

07

(d) Calculate the direction of F 

The direction of Fnet,stris leftward (radially inward).

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