Respuesta :
Answer:
Explanation:
Let l be th length of pendulum
loss of height
= mg ( l - l cos50)
= mg l ( 1-cos50)
1/2 mv² = mgl ( 1-cos50)
v = √[2gl( 1- cos50)]
= √( 2 x 9.8 x .7 x ( 1-cos50)
= 2.2 m / s
speed at the bottom = 2.2 m /s
b )
centripetal acceleration
= v² / r
= 2.2 x 2.2 / .7
= 6.9 m /s²
C )
If T be the tension
T - mg = mv² / r
T = mg + mv² / r
= .13 X 9.8 + .13 X 6.9
= 2.17 N
The speed of the ball when it reaches the bottom of the arc is 2.2 m/sec, the centripetal acceleration of the ball at this point is 6.91 m/[tex]\rm sec^2[/tex], and the tension in the string at this point is 2.1723 N.
Given :
- A small 0.13 kg metal ball is tied to a very light (essentially massless) string that is 0.70 m long.
- The string is attached to the ceiling so as to form a pendulum. The pendulum is set into motion by releasing it from rest at an angle of 50 ∘ with the vertical.
A) First, determine the distance covered by the pendulum that is 'h'.
[tex]\rm h = mg(l - l\;cos\theta )[/tex]
[tex]\rm h = mgl(1-cos\theta )[/tex]
Now, from kinematics the displacement is given by:
[tex]\rm S = \dfrac{1}{2}mv^2[/tex]
[tex]\rm mgl(1-cos\theta) = \dfrac{1}{2}mv^2[/tex]
[tex]\rm v = \sqrt{2gl(1-cos\theta )}[/tex]
Now, substitute the value of known terms in the above equation.
[tex]\rm v = \sqrt{2\times 9.8 \times 0.7(1-cos50 )}[/tex]
v = 2.2 m/sec
So, the speed of the pendulum at the bottom is 2.2m/sec.
B) The formula of the centripetal acceleration is given by:
[tex]\rm a = \dfrac{v^2}{r}[/tex]
[tex]a = \dfrac{(2.2)^2}{0.7}[/tex]
[tex]\rm a = 6.91m/sec^2[/tex]
C) Let 'T' be the tension then:
[tex]\rm T-mg = \dfrac{mv^2}{r}[/tex]
[tex]\rm T = 0.13\times 9.8+0.13\times 6.91[/tex]
T = 1.274 + 0.8983
T = 2.1723 N
For more information, refer to the link given below:
https://brainly.com/question/11048442
