An object of mass 5kg is whirled round a horizontal circle of radius 5m with uniform speed of 5m/s by a revolving string inclined to the verticat. Calculate the tension in the string and the angle of inclination of the string to the vertical.Pls show the workings
T sinθ = mg
T cosθ = mv2/r
Divide the first by the second to get
Tanθ = gr/v2
Solve for θ and use it to find T
Am a student
To calculate the tension in the string and the angle of inclination of the string to the vertical, we can analyze the forces acting on the object.
Let's consider the forces acting on the object in the horizontal circular motion:
1. Centripetal Force: The tension in the string provides the centripetal force keeping the object in circular motion. Let's denote this force as T.
2. Centrifugal Force: An outward force that tries to pull the object away from the center of the circle due to its circular motion. It is counteracted by the tension in the string.
3. Weight: The gravitational force acting vertically downwards. Let's denote the mass of the object as m (5kg) and its acceleration due to gravity as g (9.8 m/s²). The weight can be calculated as W = m * g.
Now, let's break down the weight force into its components. Since the string is inclined to the vertical, it can be resolved into two components:
1. Vertical component: This component is equal to W * sin(θ), where θ is the angle of inclination of the string to the vertical.
2. Horizontal component: This component is equal to W * cos(θ).
The tension in the string (T) must balance the vertical component of the weight, so we have:
T = W * sin(θ)
Since W = m * g, we can substitute this into the equation:
T = m * g * sin(θ)
In addition, the tension in the string (T) also provides the centripetal force required for circular motion. The centripetal force is given by:
T = (m * v²) / r
where v is the velocity (5 m/s) and r is the radius (5 m).
Now we have two equations involving T:
1. T = m * g * sin(θ)
2. T = (m * v²) / r
Since both equations represent T, we can set them equal to each other and solve for θ:
m * g * sin(θ) = (m * v²) / r
Simplifying, we have:
g * sin(θ) = (v²) / r
Now we can solve for θ:
θ = arcsin((v²) / (g * r))
Let's plug in the given values:
v = 5 m/s
r = 5 m
g = 9.8 m/s²
θ = arcsin((5²) / (9.8 * 5))
Calculating this, we find:
θ ≈ 44.4°
Now, let's calculate the tension in the string (T):
T = m * g * sin(θ)
T = 5 kg * 9.8 m/s² * sin(44.4°)
T ≈ 33.0 N
Therefore, the tension in the string is approximately 33.0 N, and the angle of inclination of the string to the vertical is approximately 44.4 degrees.
To find the tension in the string and the angle of inclination of the string to the vertical, we can use the principles of circular motion.
First, let's analyze the forces acting on the object:
1. Tension in the string: This force acts towards the center of the circle and provides the centripetal force necessary to keep the object moving in a circular path. Let's denote this force as T.
2. Gravitational force: This force acts vertically downward and is given by the formula F = mg, where m is the mass of the object (5kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
To solve for the tension in the string, we can equate it to the centripetal force required for circular motion:
T = (mass × velocity^2) / radius
Plugging in the given values, we have:
T = (5kg × (5m/s)^2) / 5m
T = 5kg × 25m^2/s^2 / 5m
T = 25kg m/s^2
Thus, the tension in the string is 25 N (newtons).
To find the angle of inclination of the string to the vertical, we can use trigonometry. Since the tension in the string acts towards the center of the circle, it can be resolved into two components: one acting vertically downward and one acting horizontally towards the center. The angle between the string and the vertical is the same as the angle between the vertical component of the tension and the downward gravitational force.
Let's denote the angle of inclination as θ. Using trigonometry, we can write:
sin(θ) = (vertical component of tension) / (gravitational force)
sin(θ) = (T) / (mg)
sin(θ) = (25N) / (5kg × 9.8m/s^2)
sin(θ) = 5 / 9.8
θ = arcsin(5 / 9.8)
Using a calculator, we can find that θ is approximately 30.5 degrees.
Therefore, the tension in the string is 25 N, and the angle of inclination of the string to the vertical is approximately 30.5 degrees.