Part A - Practice Problem: What is the period T of the ball’s motion if L =4.0m

Question

Part A - Practice Problem:

What is the period T of the ball’s motion if L=4.0m and the cord makes an angle of?=40? with the vertical?

Express your answer in seconds to two significant figures.

SOLUTION

SET UP (Figure 1) shows our diagrams. Two forces act on the ball: its weight w?  and the tension force F? T exerted by the cord. In our free-body diagram, we point the x axis toward the center of the circle (in the direction of the centripetal acceleration arad), and we replace the tension force by its x and y components: Fx=FTsin? andFy=FTcos?. The y components of force sum to zero, so the net force on the ball is the x component of tension, which acts toward the center of the ball’s circular path.

SOLVE Because the period T is known, we express arad in terms of T and the radius R of the circle, using the following equation:arad=4?2R/T2. (Note that the radius of the circular path isLsin?, not L.)

To find FT, we use Newton’s second law for the x component of net force:

?Fx=max=marad,FTsin?=m4?2RT2

Next, we substitute the expression for the radius R of the circle (R=Lsin?) into the preceding equation:

FTsin?=m4?2(Lsin?)T2

The factor sin? divides out, and we get

FT=m4?2LT2

We’re halfway home; the foregoing equation gives the rope tension FT in terms of known quantities. To obtain an expression for the angle ?, we use Newton’s second law for the y component of net force. (Remember that this component must be zero (?Fy=0)because there is no y component of acceleration.) Newton’s second law yields

?Fy=FTcos?+(?mg)=0

We substitute our expression for FT into this equation to get

m4?2LT2cos?+(?mg)=0,cos?=gT24?2L

So we’ve succeeded in expressing the tension FT and the cosine of the angle ? in terms of known quantities.

REFLECT For a given length L, cos? decreases and ? increases as T becomes smaller and the ball makes more revolutions per second. The angle can never be 90?, however; that would require that T=0, FT=?, and v=?. The relationship between T and? doesn’t involve the mass of the ball, but the force magnitude FT is directly proportional to m. When ? is very small, FT is approximately equal to the ball’s weight, cos? is approximately unity, and the expressions we found above give the approximate expressionT=2?(L/g)??????.

Part A - Practice Problem:

What is the period T of the ball’s motion if L=4.0m and the cord makes an angle of?=40? with the vertical?

Express your answer in seconds to two significant figures.

Explanation / Answer

Here ,m

for L = 4 m

angle B = 40 degree

as cos(B) = g * T^2/(4 pi^2 * L)

cos(40) = 9.8 * T^2/(4 * 3.141^2 * 4)

solving for T

T = 3.51 s

the time period us 3.51 s

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