Physics homework ??

Question

A 175 g steel ball is tied to a 2.50 m "massless" string and hung from the ceiling to make a pendulum, and then, the ball is brought to a position making a 30° angle with the vertical direction and released from rest. Ignoring the effects of the air resistance, find the speed of the ball (in m/s) when the string is at the following positions.
a) hanging vertically down
b) making an angle of 20° with the vertical
c) making an angle of 10° with the vertical

? · Accepted Answer

By geometry, the height relative to the equilibrium position for any angle Θ is
h = L*(1 - cosΘ)
so initially the ball has
GPE₀ = mgh₀ = mgL(1 - cos30º)
GPE₀ = 0.175kg*9.81m/s²*2.50m*(1 - cos30º) = 0.575 J

For any angle φ < Θ, KE = GPE₀ - mgL(1 - cosφ)

a) KE = 0.575J - 0 = ½ * 0.175kg * v² → v = 2.56 m/s ◄

b) KE = 0.575J - 0.175kg*9.81m/s²*2.50m*(1 - cos20º) = 0.316 J
0.316 J = ½ * 0.175kg * v² → v = 1.90 m/s ◄

c) KE = 0.575J - 0.175kg*9.81m/s²*2.50m*(1 - cos10º) = 0.510 J
0.510 J = ½ * 0.175kg * v² → v = 2.41 m/s ◄

Hope this helps!

Fireman · Answer

(a) The initial PE of the Bob at 30* = Total Energy of the bob = mgh = mg(L- Lcosθ) = mgL(1-cos30*) = 0.13mgL
=>E(t) = PE(30*) = 0.13 x 0.175 x 9.8 x 2.5 = 0.57 J
By he law of energy conservation
=>PE (30*) = KE(0*) = 0.57 J
=>1/2mv^2 = 0.57
=>1/2 x 0.175 x v^2 = 0.57
=>v = √6.5
=>v(at 0*) = 2.55 m/s
(b) PE(20*) = mgL(1-cos20*) = 0.175 x 9.8 x 2.5 x (1 - 0.94) = 0.26 J
Thus by the law of energy conservation,
=>0.57 = KE(20*) + 0.26
=>1/2 x 0.175 x v^2 = 0.31
=>v = √3.56
=>v(at 20*) = 1.89 m/s(c) PE(10*) = mgL(1-cos10*) = 0.175 x 9.8 x 2.5 x (1 - 0.98) = 0.07 J
Thus by the law of energy conservation,
=>0.57 = KE(20*) + 0.07
=>1/2 x 0.175 x v^2 = 0.50
=>v = √5.71
=>v(at 10*) = 2.39 m/s

Trending News

Physics homework ??

2 Answers