A small ball of mass m=0.60 kg hangs from a massless string of length l= 1.2 m. The ball travels in a vertical circle and its speed at the bottom is v0= 6.0 m/s (see figure). Neglect all friction and air drag, and use g=10 m/s2 for the gravitational acceleration. The ball is so small that we can approximate it as a point.
(a) Find the speed of the ball (in m/s) when the string is at α= 80∘.
v(α= 80∘)= 4.02889 (i got this one)
(b) What is the tension in the string (in Newton) when it is at α= 80∘?
T(α= 80∘)=
(c) The string of the pendulum is cut when it is at α= 80∘. First, we want to neglect all air drag during the trajectory of the ball. What is the maximal height h (in meters) the ball reaches above its point of release? What time tup (in s) does it take the ball to reach the highest point from the instant the string is cut? What time tdn (in s) does it take the ball to go from the highest point back to the altitude it was released from the string?
h(α= 80∘)=
tup(α= 80∘)=
tdn(α= 80∘)=
To solve these questions, we need to apply the principles of circular motion and conservation of energy.
(a) To find the speed of the ball when the string is at α= 80∘, we can use the conservation of mechanical energy. At the highest point (α= 80∘), the ball will have potential energy equal to its initial kinetic energy at the bottom of the circle. We can use the following formula to find the speed at α= 80∘:
mgh = (1/2)mv^2
Where m is the mass of the ball, g is the gravitational acceleration, h is the height difference between the bottom and α= 80∘, and v is the speed at α= 80∘.
Rearranging the formula, we get:
v = sqrt(2gh)
Substituting the given values into the formula:
v = sqrt(2 * 10 * (l - l*cos(α)))
v = sqrt(2 * 10 * (1.2 - 1.2*cos(80)))
v ≈ 4.02889 m/s
So, the speed of the ball when the string is at α= 80∘ is approximately 4.02889 m/s.
(b) To find the tension in the string at α= 80∘, we can use the centrifugal force acting on the ball. At the bottom of the circle, the tension in the string is equal to the sum of the gravitational force and the centripetal force. However, at α= 80∘, the tension is only due to the centripetal force because the gravitational force acts vertically downwards. We can use the following formula to find the tension:
T = m * V^2 / r
Where m is the mass of the ball, V is the velocity of the ball at α= 80∘, and r is the radius of the circle (equal to the length of the string).
Substituting the given values into the formula:
T = (0.60 * 4.02889^2) / 1.2
T ≈ 8.55611 N
So, the tension in the string at α= 80∘ is approximately 8.55611 N.
(c) To find the maximum height reached by the ball, we can again use conservation of energy. At the highest point, the ball will have only potential energy, which is equal to the initial mechanical energy it had at the bottom.
mgh = (1/2)mv0^2
Rearranging the formula, we get:
h = (v0^2) / (2g)
Substituting the given values into the formula:
h = (6.0^2) / (2 * 10)
h = 1.8 m
So, the maximum height reached by the ball is 1.8 m.
To find the time it takes for the ball to reach the highest point (tup), we can use the equation of motion for vertical motion:
h = v0t - (1/2)gt^2
At the highest point, the vertical displacement is zero, so the equation becomes:
0 = v0tup - (1/2)g(tup)^2
Rearranging the equation, we get a quadratic equation:
(1/2)gtup^2 - v0tup = 0
Solving this equation gives:
tup = 2v0 / g
Substituting the given values into the formula:
tup = 2 * 6.0 / 10
tup = 1.2 s
So, the time it takes for the ball to reach the highest point is 1.2 s.
To find the time it takes for the ball to go from the highest point back to the altitude it was released from the string (tdn), we can again use the equation of motion for vertical motion:
h = v0tdn - (1/2)g(tdn)^2
Substituting the given values into the formula:
1.8 = 6.0tdn - (1/2)10(tdn)^2
Simplifying the equation, we get:
5tdn^2 - 12tdn + 3.6 = 0
Solving this quadratic equation will give us tdn. By using the quadratic formula, we get:
tdn = (12 ± sqrt(12^2 - 4*5*3.6)) / (2*5)
tdn ≈ (12 ± sqrt(144 - 72)) / 10
tdn ≈ (12 ± sqrt(72)) / 10
Calculating the values for tdn, we get:
tdn ≈ (12 ± 8.48528) / 10
tdn ≈ 1.249
or
tdn ≈ 0.151
So, the ball takes approximately 1.249 s to go from the highest point back to the release altitude.
Therefore, the answers to part (c) are:
h(α= 80∘) ≈ 1.8 m
tup(α= 80∘) ≈ 1.2 s
tdn(α= 80∘) ≈ 1.249 s