A small ball of mass m=0.60 kg hangs from a massless string of length l= 0.8 m. The ball travels in a vertical circle and its speed at the bottom is v0= 5.0 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= 70∘.
(b) What is the tension in the string (in Newton) when it is at α= 70∘?
(c) The string of the pendulum is cut when it is at α= 70∘. 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= 70∘)=
tup(α= 70∘)=
tdn(α= 70∘)=
please tell the formula too
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To solve this problem, we can use the conservation of mechanical energy in combination with centripetal acceleration formulas and trigonometric functions. Let's go through each part step by step:
(a) To find the speed of the ball when the string is at α = 70∘, we can use the conservation of mechanical energy. The initial mechanical energy when the ball is at the bottom of the circle is the sum of its kinetic energy and potential energy:
E_initial = KE + PE = (1/2)mv₀^2 + mgh₀
At the top of the circle, the ball's speed reaches its minimum value, which we can denote as v_min. At this point, all of the ball's initial mechanical energy is converted to potential energy:
E_top = PE_max = mgh_max
Since kinetic energy is directly related to speed squared, we can set the initial mechanical energy equal to the mechanical energy at the top and solve for v_min:
(1/2)mv₀^2 = mgh_max
Simplifying the equation, we get:
v₀^2 = 2gh_max
Since the height at the top of the circle is given by l - h_max (the difference between the total length of the string and the maximum height reached by the ball), we can substitute this into the equation:
v₀^2 = 2g(l - h_max)
Now, we can solve for v_min:
v_min = sqrt(2g(l - h_max))
Substituting the given values, we have:
v_min = sqrt(2 * 10 * (0.8 - h_max))
(b) To find the tension in the string when it is at α = 70∘, we need to consider the forces acting on the ball at this point. The tension force provides the centripetal force necessary to keep the ball moving in a circular path. The tension force is also equal to the net force acting on the ball, which consists of the gravitational force and the centrifugal force.
At any angle α, the tension T can be expressed as:
T = mg + (mv^2 / l)
We already know the mass of the ball m and the length of the string l. To find the speed v at α = 70∘, we can use the formula for centripetal acceleration:
v = sqrt(rg)
Substituting this into the tension formula, we have:
T = mg + (mrg / l)
Substituting the given values, we get:
T = (0.60 * 10) + (0.60 * 10 * 0.8 * 5^2 / 0.8)
(c) To find the maximal height h, we need to account for the conservation of mechanical energy once the string is cut. Since we neglect air drag, the total mechanical energy of the ball remains constant throughout its trajectory.
Using the conservation of mechanical energy, the mechanical energy at the highest point is equal to the initial mechanical energy when the string is cut:
(1/2)mv² + mgh = (1/2)m(0)² + mgh_max
Simplifying the equation, we have:
v² = 2g(h_max - h)
Since we already know the initial speed v₀ and the height at the highest point h_max, we can solve for h using the equation:
v₀² = 2g(h_max - h)
To find the time tup taken by the ball to reach the highest point, we can use the fact that vertical displacement is given by:
h = h_max - (1/2)gtup²
Rearranging the equation, we get:
tup = sqrt((2(h_max - h)) / g)
To find the time tdn taken by the ball to go from the highest point back to the altitude it was released from the string, we can use the equation:
h = (1/2)gtdn²
Rearranging the equation, we get:
tdn = sqrt(2h / g)
Substituting the given values, we can find h, tup, and tdn.