A tangential force acts on the rim of a 2.0 kilogram disk-shaped wheel (0.50 m radius) and gives the wheel an angular acceleration of 4.8 rad/s^2. Neglect friction and find the magnitude of the force
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To find the magnitude of the force acting on the rim of the disk-shaped wheel, you can use the following steps:
Step 1: Determine the moment of inertia of the disk-shaped wheel. The moment of inertia (I) for a disk-shaped wheel is given by the equation:
I = (1/2) * m * r^2
where m is the mass of the wheel (2.0 kg) and r is the radius of the wheel (0.50 m). Plugging in the values, we get:
I = (1/2) * 2.0 kg * (0.50 m)^2
I = 0.5 kg.m^2
Step 2: Use Newton's second law for rotational motion, which states that the torque (τ) acting on an object is equal to the moment of inertia (I) multiplied by the angular acceleration (α).
τ = I * α
In this case, the torque (τ) is equal to the force (acting tangentially on the rim) multiplied by the radius (r):
τ = F * r
Substituting the values into the equation, we get:
F * r = I * α
Step 3: Solve for the force (F):
F = (I * α) / r
Substituting the known values, we get:
F = (0.5 kg.m^2 * 4.8 rad/s^2) / 0.50 m
Calculating:
F = 4.8 N
Therefore, the magnitude of the force acting on the rim of the disk-shaped wheel is 4.8 Newtons.
To find the magnitude of the force acting on the rim of the disk-shaped wheel, we can use Newton's second law for rotational motion.
The torque applied to the wheel is equal to the product of the force and the radius of the wheel. The torque can also be calculated as the moment of inertia multiplied by the angular acceleration.
The equation for torque is given by:
τ = I * α
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
To find the moment of inertia of a disk-shaped wheel, we can use the formula:
I = (1/2) * m * r^2
where I is the moment of inertia, m is the mass of the wheel, and r is the radius of the wheel.
Plugging in the given values:
m = 2.0 kg (mass of the wheel)
r = 0.50 m (radius of the wheel)
we can calculate the moment of inertia:
I = (1/2) * 2.0 kg * (0.50 m)^2
= 0.5 kg * 0.25 m^2
= 0.125 kg m^2
Now we can substitute the values into the torque equation:
τ = (0.125 kg m^2) * (4.8 rad/s^2)
= 0.6 Nm
Since torque is the product of force and radius, we can rewrite the equation as:
τ = F * r
Solving for the force:
F = τ / r
= 0.6 Nm / 0.50 m
= 1.2 N
Therefore, the magnitude of the force acting on the rim of the disk-shaped wheel is 1.2 Newtons.