A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.4 m/s and the tension in the rope is T = 14.9 N.
1) How long is the rope?
2) What is the mass?
3) If the maximum mass that can be used before the rope breaks is mmax = 1.2 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)
4) Now a peg is placed 4/5 of the way down the penduum.

L = 0.294 m

m = 0.507 kg

T = 35.3 N

T = 8.77 N

Explanation:

Given:

- The mass is released from rest vi = 0 m/s

- The final velocity of mass at bottom vf = 2.4 m/s

- The Tension in the rope T = 14.9 N

Find:

1) How long is the rope L?

2) What is the mass m?

3) If the maximum mass that can be used before the rope breaks is mmax = 1.2 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)

4) Now a peg is placed 4/5 of the way down the pendulum.

Solution:

- First apply the principle of conservation of energy at initial and final states as follows:

                              K.E_1 + P.E_1 = K.E_2

                              0.5*m*vi^2 + m*g*L =  0.5*m*vf^2

                              0 + m*g*L = 0.5*m*vf^2

                              L = 0.5*vf^2 / g

                              L = 0.5*2.4^2 / 9.81

                              L = 0.294 m

- Apply Newton's second Law of motion in bottom most along centripetal direction, with centripetal acceleration a_c:

                              T - m*g = m*a_c

                              a_c = vf^2 / L

                              T = m*vf^2 / L + m*g

                              m = T / (vf^2 / L + g )

                              m = (14.9)/ [ 2.4^2 / 0.294 + 9.81 ]

                              m = 0.507 kg

- For the case the maximum mass m_max required for the mass to break-off the string. Then we have the following Newton's law of motion:

                              T = m_max( vf^2 / L + g )

                              T = 1.2*(2.4^2 / 0.294 + 9.81 )

                              T = 35.3 N    

- The peg is placed at 4/5 of the path way(arc length) the angle swept is θ.

                              s = L*θ

                              4*pi*L/2*5 = L*θ

                              θ = 2*pi/5

- The vertical height h at position θ = 2*pi/5 is: Trigonometric relation

                              h = L( 1 - sin(2*pi/5))

                              h = 0.04894*L

- Apply conservation of energy and find the velocity vp when mass m hits the peg.

                             K.E_1 + P.E_1 = K.E_2 + P.E_2

                             0.5*m*vi^2 + m*g*L =  0.5*m*vp^2 + m*g*h

                             0 + m*g*L = 0.5*m*vp^2 + m*g*0.04894*L

                             vp^2 = 2*g*L*(0.95105) = sqrt (2*9.81*0.294*0.95105)

                             vp = 2.3422 m/s

- Apply the Newton's second Law of motion when mass m hits the peg.

                             T - m*g*sin(2*pi/5) = m*vp^2 / L

                             T =  m*vp^2 / L + m*g*sin(2*pi/5)

                             T = 0.507*[2.3422 / 0.294 + 9.81*sin(2*pi/5)]

                             T = 8.77 N

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

12000j is work done