One end of a cord is fixed and a small 0.300 kg object is attached to the other end, where it swings in a section of a vertical circle of radius 2.50 m as shown in the figure below. When è = 15.0°, the speed of the object is 6.20 m/s. At this instant, find each of the following.
(a) the tension in the string
T = N
(b) the tangential and radial components of acceleration
ar = m/s2 inward
at = m/s2 downward tangent to the circle
(c) the total acceleration
atotal = m/s2 inward and below the cord at °
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To solve this problem, we need to use Newton's laws of motion and some basic principles of circular motion.
(a) To find the tension in the string (T), we need to consider the forces acting on the object at that instant. These forces are the tension in the string (T) and the weight of the object (mg), where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the object is moving in a vertical circle, we can break down the forces into the radial direction (towards the center of the circle) and the tangential direction (along the tangent to the circle). At the bottom point of the swing, the tension in the string needs to provide enough force to counteract both the weight of the object as well as provide the necessary centripetal force.
The tension in the string can be found using the equation: T = mg + (mv^2)/r, where v is the speed of the object and r is the radius of the circle. Plugging in the values given in the problem:
m = 0.300 kg
v = 6.20 m/s
r = 2.50 m
g = 9.8 m/s^2
T = (0.300 kg) * (9.8 m/s^2) + (0.300 kg) * (6.20 m/s)^2 / (2.50 m)
T = 2.94 N + 4.58 N
T = 7.52 N
Therefore, the tension in the string is 7.52 N.
(b) The tangential and radial components of acceleration can be found using the following equations:
at = (v^2)/r
ar = -g
Plugging in the given values:
v = 6.20 m/s
r = 2.50 m
g = 9.8 m/s^2
at = (6.20 m/s)^2 / (2.50 m)
at = 15.24 m/s^2
ar = -9.8 m/s^2 (negative because it is in the opposite direction to the positive radial direction)
Therefore, the tangential acceleration (at) is 15.24 m/s^2 downward tangent to the circle, and the radial acceleration (ar) is 9.8 m/s^2 inward.
(c) The total acceleration (atotal) can be found by combining the tangential and radial accelerations using the Pythagorean theorem:
atotal = sqrt(at^2 + ar^2)
Plugging in the calculated values:
atotal = sqrt((15.24 m/s^2)^2 + (9.8 m/s^2)^2)
atotal = sqrt(232.66 m^2/s^4 + 96.04 m^2/s^4)
atotal = sqrt(328.70 m^2/s^4)
atotal = 18.14 m/s^2
The total acceleration (atotal) is 18.14 m/s^2 inward and below the cord at 15.0°.