Answer:
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
