You hold a bucket attached to a rope, and in the bucket is a 500.0 g rock. You swing the bucket so the rock moves in a vertical circle 1.6 m in diameter.
What is the minimum speed the rock must have at the top of the circle if it is to always stay in contact with the bottom of the bucket?
mv²/R =mg
v=sqrt(gR)
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To find the minimum speed the rock must have at the top of the circle, we can use the concept of centripetal force.
When the rock moves in a vertical circle, it experiences two forces: the gravitational force acting downward and the tension force acting inward towards the center of the circle. For the rock to always stay in contact with the bottom of the bucket, the tension force must provide the necessary centripetal force to balance the gravitational force.
The centripetal force is given by the equation:
F_c = (m * v^2) / r
where F_c is the centripetal force, m is the mass of the rock, v is the velocity of the rock, and r is the radius of the circle (which is half the diameter).
At the top of the circle, the tension force and the gravitational force are acting in opposite directions. So, the centripetal force is the difference between these two forces:
F_c = T - mg
where T is the tension force and mg is the gravitational force.
Setting these equations equal to each other:
T - mg = (m * v^2) / r
Now, we can substitute the values given in the question:
m = 500.0 g = 0.5 kg (converting grams to kilograms)
r = 1.6 m
g = 9.8 m/s^2 (acceleration due to gravity)
v is the minimum speed we want to find.
Plugging in these values:
T - (0.5 kg * 9.8 m/s^2) = (0.5 kg * v^2) / 1.6 m
Simplifying the equation:
T - 4.9 N = 0.3125 kg.m/s^2
Now, we need to find the tension force T. To do this, we need to consider the bottom of the circle, where the tension force and the gravitational force act in the same direction. At the bottom:
F_c = T + mg
Setting this equal to the centripetal force:
T + mg = (0.5 kg * v^2) / 1.6 m
Plugging in the values:
T + (0.5 kg * 9.8 m/s^2) = (0.5 kg * v^2) / 1.6 m
T + 4.9 N = 0.3125 kg.m/s^2
Now, we have a system of two equations with two unknowns (T and v). We can solve this system of equations to find the minimum speed v at the top of the circle.
Solving the system of equations:
T - 4.9 N = 0.3125 kg.m/s^2
T + 4.9 N = 0.3125 kg.m/s^2
By adding the two equations together, we can eliminate T:
2T = 0.625 kg.m/s^2
T = 0.3125 kg.m/s^2
Now, we can substitute this value of T into one of the original equations to solve for v:
0.3125 kg.m/s^2 - 4.9 N = (0.5 kg * v^2) / 1.6 m
Simplifying the equation:
-4.9 N = (0.3125 kg * v^2) / 1.6 m
Rearranging the equation:
v^2 = (-4.9 N * 1.6 m) / 0.3125 kg
Calculating the right side of the equation:
v^2 = -24.8 N.m / 0.3125 kg
v^2 = -79.36 m^2/s^2
Taking the square root of both sides:
v = √(-79.36) m/s
The minimum speed v at the top of the circle, rounded to two decimal places, is approximately 8.91 m/s.