Question

A pendulum string with the length of 105 cm has the 325 g bob attached to it. At the lowest point of the swing the pendulum bob moves with the speed of 2.15 m/s. Determine: a) the centripetal acceleration of the pendulum bob at this point; b) the force of tension in the string at this point; c) the kinetic energy of the pendulum bob at this point;

Answer 1

Answer:

A. 4.40m/s²

B. 4.615N

C. 0.7511J

Explanation:

A. Ac= V²/L

= (2.15)²/1.05

= 4.50m/s²

B. T =Mv²/L + mg

0.325(4.40)+ 0.325(9.8)

= 4.615N

C. K.E = 1/2mv²

=0.7511J

You can also refer to attached handwritten document for more details

Answer 2

Answer:

(a) 4.4 m/s²

(b) 4.615 N.

(c) 0.751 J

Explanation:

(a)

Using,

a' = v²/r..................... Equation 1

Where a' = centripetal acceleration, v = speed of the pendulum bob, r = radius or length of the pendulum bob.

Given: v = 2.15 m/s, r = 105 cm = 1.05 m

Substitute into equation 1

a' = 2.15²/1.05

a' = 4.4 m/s².

(b) The force of tension in the string = Tangential force + weight of the bob.

T = ma'+mg..................... Equation 2

Where T = Force of tension in the string, m = mass of the bob, g = acceleration due to gravity.

Given: m = 325 g = 0.325 kg, a' = 4.4 m/s², g = 9.8 m/s²

Substitute into equation 2

T = 0.325×4.4+0.3325×9.8

T = 1.43+3.185

T = 4.615 N.

(c) Kinetic energy of he bob at that point = 1/2mv²

Ek = 1/2mv²...................... Equation 3

Where Ek = kinetic energy of the bob

Given: m = 0.325 kg, v = 2.15 m/s

Substitute into equation 3

Ek = 1/2(0.325)(2.15²)

Ek = 0.751 J