Question

A simple pendulum consists of a small object of mass m= 0.150 kg suspended from a support stand by a light string. The string has a length L= 0.750 m. The string has an initial position given by θ= 65.0° relative to the vertical. The pendulum is released from rest. Air resistance is negligible during the subsequent motion of the pendulum.

a)Calculate the work done by gravity on the pendulum as it moves from its initial position to the lowest point of its semicircular arc. b)Calculate the speed of the object as it moves through the lowest point of its semicircular arc.

c)Calculate the tension in the string right as the object is moving through the lowest point of its semicircular arc.

d)Repeat part b) using the conservation of total mechanical energy. (Is the total mechanical energy of the pendulum conserved?)

Answer 1

a) given mass of the pendulum M = 0.150 kg length of the string L =0.750 m    = 65 degrees

W= Work done is the change in potential energy = MgH2-MgH1

   H1 = L( 1 - cos ) H2 =0

   W = Mg(H2-H1) = 0.15 (9.8)(0-(0.750(1-cos65)))

W =- 0.636 J

b)    Change in kinetic energy is equal to change in potential energy

W = -(1/2)M(V2^2-V1^2)

   but V1= 0 at initial position   

   0.636 = 0.5 (0.150)(V2^2)

   V2^2 = 8.48

   V2 = 2.91m/s

c) Tension in the string will be equal to the sum of centrigular force and weight

   centrifugal force = mV2^2/R   

Tension T = (M(V2^2)/r)+Mg

T = ((0.15(2.91)^2)/0.750) + 0.15(9.8)

T =3.16 N

d) Total energy of the pendulum is always conserved .

at initial position H1 = L(1-cos ) =0.750(1-cos65) =0.433    V1 =0

total energy = MgH1+0.5 M V1^2

   = 0.15(9.8)(0.433)

   T.E1 = 0.636 J

   Total energy at lowest point TE2 = MgH2 +0.5MV2^2

   H2 =0    V2 = 2.91 m/s

   TE = 0.15(9.8)(0)+(0.5)(0.15)(2.91)^2

   TE2 = 0.636 J

TE1 = TE2 therefore total energy of the pendulum is always conserved.