A force of 200 N acts tangentially on the rim of a wheel 25 cm in radius. (a) Find the torque. (b) Repeat if
the force makes an angle of 40 to a spoke of the wheel.
(a) The torque is given by:
τ = F × r
where F is the force and r is the radius of the wheel. Substituting the given values, we get:
τ = 200 N × 0.25 m
τ = 50 Nm
Therefore, the torque is 50 Nm.
(b) When the force makes an angle of 40 to a spoke of the wheel, the torque is given by:
τ = F × r × sinθ
where θ is the angle between the force and the radius of the wheel. Substituting the given values, we get:
τ = 200 N × 0.25 m × sin40
τ ≈ 77.2 Nm
Therefore, the torque is approximately 77.2 Nm.
A certain 8.0-kg wheel has a radius of gyration of 25 cm. (a) What is its moment of inertia? (6) How large a
torque is required to give it an angular acceleration of 3.0 rad/s?
(a) The moment of inertia of a wheel rotating about its axis of symmetry is given by:
I = k^2 × M
where k is the radius of gyration and M is the mass of the wheel. Substituting the given values, we get:
I = (0.25 m)^2 × 8.0 kg
I = 0.50 kg m^2
Therefore, the moment of inertia of the wheel is 0.50 kg m^2.
(b) The torque required to give the wheel an angular acceleration of 3.0 rad/s^2 can be calculated using the equation:
τ = I × α
where τ is the torque, I is the moment of inertia, and α is the angular acceleration. Substituting the given values, we get:
τ = 0.50 kg m^2 × 3.0 rad/s^2
τ = 1.5 Nm
Therefore, a torque of 1.5 Nm is required to give the wheel an angular acceleration of 3.0 rad/s^2.
Determine the constant torque that must be applied to a 50-kg flywheel, with radius of gyration 40 cm, to give it a frequency of 300 rpm in 10 s if it's initially at rest.
The frequency, f, of a rotating object is given by:
f = (1/2π) × (ω)
where ω is the angular velocity of the object. We can convert the given frequency of 300 rpm to radians per second:
ω = 2π × 300 rpm / 60
ω = 31.42 rad/s
The moment of inertia of the flywheel is given by:
I = k^2 × M
where k is the radius of gyration and M is the mass of the flywheel. Substituting the given values, we get:
I = (0.40 m)^2 × 50 kg
I = 8.0 kg m^2
To determine the required torque, we can use the rotational version of Newton's second law:
τ = I × α
where τ is the torque, I is the moment of inertia, and α is the angular acceleration. From the given information, we know that we want to reach an angular velocity of 31.42 rad/s in 10 seconds, so the angular acceleration required is:
α = ω / t
α = 31.42 rad/s / 10 s
α = 3.142 rad/s^2
Substituting the known values into the torque equation, we get:
τ = 8.0 kg m^2 × 3.142 rad/s^2
τ = 25.13 Nm
Therefore, a constant torque of 25.13 Nm must be applied to the flywheel to give it a frequency of 300 rpm in 10 s if it's initially at rest.
A 4.0-kg wheel of 20 cm radius of gyration is rotating at 360 rpm. The retarding frictional torque is 0.12 Nm. Compute the time it would take to coast to rest
We can start by calculating the initial angular velocity of the wheel, which is:
ω = 2π × 360 rpm / 60
ω = 37.7 rad/s
The moment of inertia of the wheel is given by:
I = k^2 × M
where k is the radius of gyration and M is the mass of the wheel. Substituting the given values, we get:
I = (0.20 m)^2 × 4.0 kg
I = 0.16 kg m^2
The retarding frictional torque can be written as:
τ = -ηω
where η is the frictional coefficient and τ is the torque. In this case, τ is given to be -0.12 Nm, so we can solve for η:
η = τ / ω
η = -0.12 Nm / 37.7 rad/s
η ≈ -0.00319 Nm s/rad
The negative sign of η indicates that it acts in the opposite direction to the initial rotation of the wheel, which is why it is retarding.
Using the equation of rotational motion:
ω = ω0 + αt
where ω0 is the initial angular velocity, α is the angular acceleration, and t is the time, we can solve for the time it takes for the wheel to come to rest. At rest, the final angular velocity is zero, so we have:
0 = ω0 + αt
Solving for t, we get:
t = -ω0 / α
The angular acceleration, α, can be determined from the retarding frictional torque and the moment of inertia:
τ = Iα
Solving for α and substituting the known values, we get:
α = τ / I
α = -0.12 Nm / 0.16 kg m^2
α = -0.75 rad/s^2
Substituting the calculated values of ω0 and α into the equation for time, we get:
t = -37.7 rad/s / (-0.75 rad/s^2)
t = 50.3 s
Therefore, it would take 50.3 seconds for the 4.0-kg wheel with radius of gyration 20 cm to coast to rest under the influence of a retarding frictional torque of 0.12 Nm.
How large a force is required to accelerate a 1300-kg car from rest to a speed of 20 m/s in a distance of
80 m?
We can use the equations of kinematics and dynamics to solve this problem.
First, we can determine the acceleration of the car using the kinematic equation:
v^2 = v0^2 + 2aΔx
where v is the final velocity, v0 is the initial velocity (zero in this case), a is the acceleration, and Δx is the distance traveled. Substituting the given values, we get:
(20 m/s)^2 = 0 + 2a(80 m)
a = 5 m/s^2
Now that we know the acceleration, we can use Newton's second law of motion to determine the force required:
F = ma
where F is the force, m is the mass of the car, and a is the acceleration calculated above. Substituting the given values, we get:
F = 1300 kg × 5 m/s^2
F = 6500 N
Therefore, a force of 6500 N (or approximately 650 kg) is required to accelerate the 1300-kg car from rest to a speed of 20 m/s in a distance of 80 m.
The following coplanar forces pull on a ring: 200 N at 30.0°, 500 N at 80.0°, 300 N at 240°, and an unknown
force. Find the magnitude and direction of the unknown force if the ring is to be in equilibrium.