Question

The pendulum shown to the left consists of a bob of mass of 0.44 kg, attached to a string of length L = 1.2 meters. It is moving to the right with a speed of 4.62 m/sec at point A. What we'd like to know is what the the velocity of the bob when it reaches point B which is 36 degrees from vertical.

There are two different ways to look at this problem, each of which will give us the same answer. Let's look at each separately. (For all of the questions below, let's start by defining the potential energy of the bob to be zero at point A.)

First, lets look at the question in terms of the work energy theorem. (W_TOT=?KE)

1)

What is the initial kinetic energy of the bob at point A?

KE_A =

J

2)

What is the work done by on the bob by gravity as it moves from point A to point B?

W_GRAV =

J  

3)

What is the work done by on the bob by the force of tension from the string attached to the bob as it moves from point A to point B?

W_TEN =

J  

4)

What is the total work done on the bob as it moves from point A to point B?

W_TOT =

J

5)

What is change in kinetic energy of the bob as it moves from point A to point B?

?KE =

J

6)

What is kinetic energy of the bob at point B?

KE_B =

J

7)

What is velocity of the bob at point B?

v_B =

m/sec

8)

Now, lets look at the problem in terms of mechanical energy. That is,ME=U+KE and ?ME=W_non-con. (Recall we aredefining the potential energy to be zero at point A.) What is the kinetic energy of the bob at point A?

KE_A =

J

9)

What is the potential energy of the bob at point A?

U_A=

J

10)

What is the mechanical energy of the bob at point A?

ME_TOT-A=

J

11)

What is work done by non-conservative forces as the bob moves from point A to point B?

W_non-con=

J

12)

What is the mechanical energy of the bob at point B?

ME_TOT-B=

J

13)

What is the potential energy of the bob when it reaches point B?

U_B =

J

14)

What is the kinetic energy of the bob when it reaches point B?

KE_B =

J

15)

What is the speed of the bob at point B?

v=

Answer 1

PART (A)

(1) KE_A = 1/2 * M * V_A² = 0.5 * 0.44 * 4.62² = 4.70 joule

(2) W_g = -Mg ( L - Lcos36) = - 0.44 * 9.8 * 1.2 ( 1- cos36) = - 0.988 joule

(3) W_tension = 0 , becaz tension is perpendicular to displacement everytime.

(4) W_total = W_g + W_tension = - 0.988 joule

(5) change in KE = total work done = - 0.988 joule

(6) KEb = - MgL( 1 - cos36) + 1/2 * M * V_a² = - 0.988 + 4.70 = 3.71 joule

(7) V_b = [ 2 * KEb / M ]^1/2 = 4.106 m/s

(8) KE_A = 4.70 joule

(9) PE_A = 0

(10) MEa = KEa + PEa = 4.70 joule

(11) W_non-con = 0

(12) ME_B = 4.70 joule

(13) PE_b = mgL ( 1- cos36) = 0.988 joule

(14) KEb = MEb - PEb = 4.70 - 0.988 =3.71 joule

(15) Vb = 4.106 m/s