A child exerts a tangential 43.5-N force on the rim of a disk-shaped merry-go-round with a radius of 2.41 m. If the merry-go-round starts at rest and acquires an angular speed of 0.086 rev/s in 3.54 s, what is its mass?
Applied torque = 43.5 * 2.41 = 104.84 Newton-meters
Angular acceleration rate =
0.086*2*pi/3.54 = 0.1526 rad/s^2
The moment of inerti, I , is torque divided by angular acceleration.
I = 687 kg*m^2
For a uniform disc, The moment of inertia is
I = (1/2) M R^2.
Use that equation and the known value of I so solve for M.
Well, to figure out the mass of the merry-go-round, we need to consider a few things. First, the child exerts a tangential force on the rim of the merry-go-round, which helps in accelerating it. We can use the formula for torque, which is force multiplied by the radius, to calculate the torque exerted by the child.
Now, for the mass of the merry-go-round, we know that it starts at rest and acquires an angular speed over a certain time. This means that there is an acceleration occurring, which is caused by the torque from the child's force. We can use Newton's second law for rotation, which states that torque is equal to the moment of inertia multiplied by the angular acceleration.
Since we don't know the moment of inertia, we need to rearrange the formula to isolate it:
Moment of inertia = Torque / Angular Acceleration
Now, we can calculate the torque exerted by the child using the force and radius given. The angular acceleration can be found by dividing the change in angular speed (0.086 rev/s) by the time (3.54 s).
Once we have both torque and angular acceleration, we can plug them into the formula to find the moment of inertia. Finally, the mass can be calculated using the formula:
Mass = Moment of inertia / Radius²
So, it's like trying to solve a puzzle. Good luck, and remember, no clowns were harmed in the making of this response!
To solve this problem, we can use the principle of conservation of angular momentum. The initial angular momentum of the merry-go-round is zero since it starts at rest. The final angular momentum can be calculated using the given information.
Angular momentum (L) is given by the formula:
L = moment of inertia (I) × angular velocity (ω)
The moment of inertia of a disk-shaped object is given by:
I = (1/2) × mass (M) × radius^2
Given:
Tangential force (F) = 43.5 N
Radius (r) = 2.41 m
Angular speed (ω) = 0.086 rev/s
Time (t) = 3.54 s
Step 1: Convert the angular speed from rev/s to rad/s:
Since 1 revolution (rev) is equal to 2π radians, we can convert rev/s to rad/s by multiplying by 2π.
ω = 0.086 rev/s × 2π rad/rev
ω = 0.086 × 2π rad/s
ω ≈ 0.54 rad/s
Step 2: Calculate the final angular momentum:
L = I × ω
We need to find I first using the formula above.
I = (1/2) × M × r^2
Step 3: Substitute the values into the formulas and solve for I:
0.54 = (1/2) × M × (2.41)^2
Step 4: Solve for M:
Multiply both sides by 2:
1.08 = M × (2.41)^2
Divide both sides by (2.41)^2:
M = 1.08 / (2.41)^2
M ≈ 0.193 kg
Therefore, the mass of the merry-go-round is approximately 0.193 kg.
To find the mass of the merry-go-round, we can use the concept of torque and rotational motion. The torque exerted on the merry-go-round is equal to the product of the force and the lever arm.
The torque can be calculated using the formula:
Torque = Force × Lever arm
Given that the tangential force exerted by the child is 43.5 N and the radius of the merry-go-round is 2.41 m, we can calculate the torque:
Torque = 43.5 N × 2.41 m
Now, let's calculate the torque:
Torque = 104.835 N·m
The torque is also equal to the product of the moment of inertia (I) and the angular acceleration (α) of the merry-go-round:
Torque = I × α
We are given the angular speed (ω) of the merry-go-round, which is 0.086 rev/s. To convert this to radians per second, we need to multiply it by 2π:
ω = 0.086 rev/s × 2π rad/rev
ω = 0.172π rad/s
We are also given the time (t) taken for the merry-go-round to acquire this angular speed, which is 3.54 s.
The angular acceleration (α) can be calculated using the following formula:
α = ω / t
Let's calculate α:
α = (0.172π rad/s) / 3.54 s
Now, we have the torque and the angular acceleration. By setting them equal to each other, we can find the moment of inertia (I).
I × α = Torque
I = Torque / α
Plug in the values:
I = 104.835 N·m / [(0.172π rad/s) / 3.54 s]
Now, we can solve for the moment of inertia (I).
I = 104.835 N·m / (0.172π rad/s) × (3.54 s)
I ≈ 196.98 kg·m²
The moment of inertia (I) of the merry-go-round represents the mass distribution of the object. It depends on the shape and mass distribution of the merry-go-round. However, in this case, we assume that the moment of inertia is proportional to the mass of the object.
Therefore, we can conclude that the mass of the merry-go-round is approximately 196.98 kg.