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∘)=
the rest i got them right :)
To find the tension in the string when it is at α= 80∘, we can use the concept of centripetal force.
(b) The tension in the string is equal to the centripetal force acting on the ball. At any point in the circular motion, the net force is given by the difference between the weight of the ball and the tension in the string.
At the bottom of the circle (α=0), the tension is equal to the weight of the ball plus the centripetal force:
T(α=0) = mg + m*v0^2/l
In this case, T(α=0) = 0.60 kg * 10 m/s^2 + 0.60 kg * (6 m/s)^2 / 1.2 m = 6 N + 18 N = 24 N.
To find the tension at α= 80∘, we need to consider the component of gravity that acts tangentially to the circle and subtract it from the total tension at α= 0∘.
At α= 80∘, the component of gravity tangential to the circle is given by the equation:
Fg_tan = m * g * sin(α)
T(α=80∘) = T(α=0) - Fg_tan
Substituting the values:
T(α=80∘) = 24 N - (0.60 kg * 10 m/s^2 * sin(80∘))
Using this formula, you can calculate the tension in the string when α= 80∘.
(c) To find the maximal height h the ball reaches above its point of release, we can use the conservation of mechanical energy.
Since there is no air drag, the total mechanical energy of the ball is conserved throughout its motion. At the point of release, all the energy is in the form of kinetic energy. At the highest point, the ball has reached its maximum height, and all the energy is in the form of potential energy.
Using the conservation of mechanical energy equation:
KE_initial + PE_initial = KE_final + PE_final
At the point of release, there is no potential energy, so the equation becomes:
KE_initial = KE_final + PE_final
The initial kinetic energy is given by:
KE_initial = 0.5 * m * v0^2
At the highest point, the final kinetic energy is zero because the ball comes to rest momentarily, and the final potential energy is given by:
PE_final = m * g * h
Setting the initial kinetic energy equal to the sum of the final potential energy and final kinetic energy, we can solve for h:
0.5 * m * v0^2 = m * g * h
h = 0.5 * v0^2 / g
Using this formula, you can calculate the maximal height h the ball reaches above its point of release.
To find the time tup the ball takes to reach the highest point from the instant the string is cut, we can use the equation for the time of flight in projectile motion.
Since the ball is released at an angle and not purely vertically, we can use the initial velocity component along the vertical direction to calculate the time of flight. The initial vertical component of the velocity is given by:
v0_y = v0 * cos(α)
The time of flight is then given by the formula:
tup = 2 * v0_y / g
Using this formula, you can calculate the time tup it takes for the ball to reach the highest point from the instant the string is cut.
To find the time tdn it takes for the ball to go from the highest point back to the altitude it was released from the string, we can use the equation for the time of flight in projectile motion.
Since the ball is in freefall, the time to reach the ground is equal to the time it takes to reach the highest point. Therefore, tdn is also equal to tup.
Using these equations, you can calculate the time tdn it takes for the ball to go from the highest point back to the altitude it was released from the string.