A ball (mass m = 250 g) on the end of an ideal string is moving in a circular motion as a conical pendulum. The length L of the string is 1.84 m and the angle with the vertical is 37 degrees.
a) What is the magnitude of the torque (N m) exerted on the ball about the support point?
b) What is the magnitude of the angular momentum (kg m^2/2) of the ball about the support point?
Correct Answers: a) 2.71 b) 1.32
For a I assumed 0 because there wasn't any said force. I do not know how to solve problem.
For b I used L = m*v*r where L = momentum
m = .250 kg
v = (r * g * tan 37 )^(0.5) = (L*sin 37*9.8*tan 37)^(0.5) = 2.8596
r = L * sin 37 = 1.84 * sin 37= 1.10733
Therefore L = .791 This was supposedly incorrect
a. torque= LXforce, force is mg downward.
torque= 1.84*.250*9.8*sin(180-37)
Any idea how I was hung up on 2nd one?
To solve this problem, we can start by analyzing the forces acting on the ball in motion.
a) Torque:
In circular motion, a torque is exerted on an object due to the force acting perpendicular to the radius of the circular path. In this scenario, the force that causes the circular motion is the tension in the string.
To calculate the torque, we can use the equation:
Torque = force * radius
In this case, the tension force is equal to the weight of the ball, since the ball is in equilibrium. The weight force is given by:
Force = mass * acceleration due to gravity
Substituting the given values,
Force = 0.250 kg * 9.8 m/s^2 = 2.45 N
The radius of the circular path is given by L * sin(angle). Substituting the given values,
Radius = 1.84 m * sin(37 degrees) = 1.11 m
Now, we can calculate the torque:
Torque = 2.45 N * 1.11 m = 2.71 N m
Therefore, the magnitude of the torque exerted on the ball about the support point is 2.71 N m.
b) Angular Momentum:
The angular momentum of an object in circular motion is given by the product of its moment of inertia and its angular velocity.
To calculate the angular momentum, we can use the equation:
Angular Momentum = moment of inertia * angular velocity
The moment of inertia of a point mass rotating about an axis perpendicular to its motion (like in this case) is given by:
Moment of Inertia = mass * radius^2
Substituting the given values,
Moment of Inertia = 0.250 kg * (1.11 m)^2 = 0.30975 kg m^2
The angular velocity can be calculated by dividing the linear velocity by the radius:
Angular Velocity = linear velocity / radius
The linear velocity can be calculated using the equation v = r * ω, where ω is the angular velocity.
Substituting the given values and rearranging the equation, we get:
Angular Velocity = v / r = (L * sin(angle) * g * tan(angle))^(0.5) / (L * sin(angle))
Substituting the given values,
Angular Velocity = (1.84 m * sin(37 degrees) * 9.8 m/s^2 * tan(37 degrees))^(0.5) / (1.84 m * sin(37 degrees))
Angular Velocity ≈ 4.95 rad/s
Now, we can calculate the angular momentum:
Angular Momentum = 0.30975 kg m^2 * 4.95 rad/s ≈ 1.53 kg m^2
However, this value is not the same as the correct answer provided (1.32 kg m^2/2).
Please recheck the given values and equations used for calculating the angular momentum.
(a) There two forces acting on the ball: gravity and tension. The line of tension psses through the support point => its torque about this point is zero.
The totque of gravity about support point is
τ=mg•Lsinα=0.25•9.8•1.84•sin37º= 2.71 N•m
(b)
x- and y-projections of acting forces are
x: m•v²/R= T•sinα, (1)
y: m•g=T•cosα. (2)
Divide (1) by (2)
m•v²/R•m•g = T•sinα/ T•cosα
v=sqrt(R•g•tanα) =
=sqrt(L•sinα• g•tanα)=
sqrt(1.84•9.8•0.6•0.75) =2.86 m/s.
Angular momentum =m•v•L=
0.25•2.86•1.84=1.32 kg•m²/s.