A bicycle wheel has a diameter of 47.6 cm and a mass of 0.809 kg. The bicycle is placed on a stationary stand on rollers and a resistive force of 60.1 N is applied to the rim of the tire. Assume all the mass of the wheel is concentrated on the outside radius.
In order to give the wheel an acceleration of 3.18 rad/s2, what force must be applied by a chain passing over a sprocket with diameter 4.16 cm?
Answer in units of N.
Moment of inertia of the wheel is I=MR^2
From resistance force=-force*R
net torque= momentInertia*angular acceleration
a. figure the moment of inertia for the wheel.
b. Net torque= force*.0416/2-60.1*.476/2
c. you are given angular acceleration
figure then, force.
I ask you and the response is asking sorry I will never use this application.
The moment of inertia of the bicycle wheel can be calculated using the formula I = MR^2, where M is the mass of the wheel and R is the radius of the wheel.
Given that the diameter of the wheel is 47.6 cm, the radius would be half of that, which is 23.8 cm (0.238 m). The mass of the wheel is given as 0.809 kg.
So, the moment of inertia (I) of the wheel is:
I = (0.809 kg) * (0.238 m)^2
Next, let's calculate the torque exerted by the resistive force.
The torque exerted by the resistive force is given by the formula torque = force * radius.
In this case, the force is given as 60.1 N, and the radius is the same as before, 0.238 m.
So, the torque (τ) exerted by the resistive force is:
τ = 60.1 N * 0.238 m
According to Newton's second law for rotational motion, the torque (τ) is equal to the moment of inertia (I) times the angular acceleration (α).
So, τ = I * α.
We want to find the force that must be applied by a chain passing over a sprocket with a diameter of 4.16 cm (0.0416 m) to give the wheel an acceleration of 3.18 rad/s^2. This force will produce the same torque as the resistive force.
Let's solve for the force:
τ = I * α
60.1 N * 0.238 m = (0.809 kg) * (0.238 m)^2 * 3.18 rad/s^2
Dividing both sides of the equation by 0.0416 m:
Force = (0.809 kg) * (0.238 m) * 3.18 rad/s^2 / 0.0416 m
Now let's calculate the force:
Force = (0.809 kg) * (0.238 m) * 3.18 rad/s^2 / 0.0416 m
The force required by the chain passing over the sprocket is approximately 14.63 N (rounded to two decimal places).
To find the force that must be applied by the chain passing over the sprocket, we need to use Newton's second law, which states that the force applied is equal to the mass multiplied by the acceleration. In this case, the mass refers to the mass of the wheel.
First, let's calculate the moment of inertia of the wheel (I) using the formula I = MR^2, where M is the mass of the wheel, and R is the radius of the wheel.
Given that the diameter of the wheel is 47.6 cm, the radius (R) can be calculated by dividing the diameter by 2:
R = 47.6 cm / 2 = 23.8 cm = 0.238 m
The mass of the wheel is given as 0.809 kg.
So, I = 0.809 kg * (0.238 m)^2 = 0.0459 kg·m^2
Next, recall the equation for torque (τ) given by τ = force × radius. In this case, the force is the force applied by the chain, and the radius is half of the sprocket's diameter.
The diameter of the sprocket is given as 4.16 cm, so the radius (r) can be calculated as follows:
r = 4.16 cm / 2 = 2.08 cm = 0.0208 m
Now, we have τ = force × 0.0208 m.
The resistive force applied to the wheel is 60.1 N, and it acts at the wheel's rim. From the equation τ = force × radius, we can write:
60.1 N = force × 0.0208 m
Rearranging the equation to solve for force, we get:
force = 60.1 N / 0.0208 m
Evaluating this expression, we find that the force required is approximately 2884.61 N. Therefore, to give the wheel an acceleration of 3.18 rad/s^2, a force of 2884.61 N must be applied by the chain passing over the sprocket.