The axle of a solid wheel of mass M and radius R is attached to a horizontal spring of constant k as shown below. Derive and simplify an expression for the period of the wheel’s oscillation, assuming it rolls back and forth without slipping.
Answer
T= 2pi squrt(3M/2k)
To derive the expression for the period of oscillation of the wheel, we can use the concept of rotational motion and the properties of a torsional spring.
First, let's understand the forces acting on the wheel. When the wheel is oscillating, it experiences two main forces: the gravitational force (mg) and the restoring force from the spring (F). The gravitational force can be assumed to act at the center of mass of the wheel, and the restoring force from the spring acts at the point where the axle is attached to the spring.
The net force acting on the wheel is the difference between these two forces:
Net Force = F - mg
Now, let's analyze the rotational motion of the wheel. Since the wheel rolls back and forth without slipping, the torque exerted by the net force will cause angular acceleration. The torque is the product of the net force and the radius of the wheel (R):
Torque = (F - mg) * R
According to Newton's second law for rotational motion, the torque is equal to the moment of inertia (I) multiplied by the angular acceleration (α):
Torque = I * α
For a uniform solid wheel, the moment of inertia about the axle is given by:
I = (1/2) * M * R^2
Now, let's equate the two expressions for torque:
(F - mg) * R = (1/2) * M * R^2 * α
Since the angular acceleration (α) is related to the angular displacement (θ) and period of oscillation (T) as α = (2π / T)^2 * θ, we can substitute this into the equation:
(F - mg) * R = (1/2) * M * R^2 * [(2π / T)^2 * θ]
Simplifying the equation, we have:
(F - mg) = (1/2) * M * R * (2π / T)^2 * θ
Since the force from the spring (F) is directly proportional to the angular displacement (θ) and its proportionality constant is the torsional spring constant (k):
F = -k * θ
Substituting this into the equation, we get:
-k * θ - mg = (1/2) * M * R * (2π / T)^2 * θ
Rearranging the terms, we obtain:
- mg = (1/2) * M * R * (2π / T)^2 * θ - k * θ
Factoring out θ, we get:
- mg = θ * [(1/2) * M * R (2π / T)^2 - k]
Now, we can solve for the angular displacement (θ):
θ = -mg / [(1/2) * M * R (2π / T)^2 - k]
Since the time period T is related to the angular displacement (θ) and angular velocity (ω) as T = (2π / ω), we can substitute this into the equation:
θ = -mg / [(1/2) * M * R (4π^2 / T^2) - k]
Simplifying further, we have:
θ = -mg / [(2M / T^2) * R - k]
Finally, we can find the value of T by solving this equation for θ = 2π:
2π = -mg / [(2M / T^2) * R - k]
Rearranging the terms and simplifying, we get:
T^2 = (4π^2 * M * R - 2π * mg) / k
Taking the square root of both sides, we obtain the expression for the period of oscillation (T):
T = 2π √[(M * R - mg/2) / k]
Finally, we can further simplify the expression by noting that mg/2 is the weight of the wheel, which is equal to the spring force at equilibrium (F₀), so the equation becomes:
T = 2π √[(M * R - F₀) / k]
Therefore, the simplified expression for the period of oscillation of the wheel is:
T = 2π √[(M * R - F₀) / k]