A 0.2 kg yo-yo consists of two solid disks of radius 5.5 cm joined together by a massless rod of radius 1 cm and a string wrapped around the rod. One end of the string is held fixed and is under constant tension T as the yo-yo is released.
a. Find the acceleration of the yo-yo.
b. Find the tension T.
Given:
m=0.2
R=5.5
r=1
l = 0.5*mR^2
=(0.5*0.2*5.5^2)
=3.025
a. a=g[1 - I/(mr^2 + I)]
=9.8[1 - (0.5*0.2*5.5^2)/(0.2*1^2 + (0.5*0.2*5.5^2))]
=0.61 m/s^2
b. T=g/[r^2/I + 1/m]
=9.8/[1^2/(0.5*0.2*5.5^2) + 1/0.2]
=1.84 N
To find the acceleration of the yo-yo and the tension T, we need to analyze the forces acting on the yo-yo and apply Newton's second law of motion.
a. Find the acceleration of the yo-yo:
The only external force acting on the yo-yo is the tension T in the string. The other forces acting on the yo-yo, such as the gravitational force and the normal force, do not contribute to the acceleration of the center of mass because they act on the rotational motion.
We can write the equation for the net force acting on the yo-yo as:
F_net = T - mg = ma
Where m is the mass of the yo-yo (0.2 kg), g is the acceleration due to gravity (9.8 m/s^2), and a is the acceleration of the yo-yo (unknown).
Rearranging the equation and solving for a, we get:
a = (T - mg) / m
b. Find the tension T:
To find the tension, we need to consider the rotational motion of the yo-yo. The yo-yo experiences a torque due to the tension force acting at a distance from the center of mass.
The torque (τ) can be calculated as the product of the tension force (T) and the moment arm (r), which is the perpendicular distance from the center of mass to the point where the tension force is applied. In this case, the moment arm is equal to the radius of the rod (r = 1 cm = 0.01 m).
The torque can be written as:
τ = T * r
The torque τ is also equal to the moment of inertia (I) multiplied by the angular acceleration (α). For a solid disk rotating about an axis perpendicular to its plane, the moment of inertia is given by:
I = (1/2) * m * r^2
Where m is the mass of the disk (0.2 kg) and r is the radius of the disk (5.5 cm = 0.055 m).
The angular acceleration α is related to the linear acceleration a by the equation:
α = a / r
Substituting the expressions for torque, moment of inertia, and angular acceleration, we get:
T * r = (1/2) * m * r^2 * (a / r)
Simplifying the equation, we find:
T = (1/2) * m * r * a
Now we can substitute the expression for a that was obtained in part a:
T = (1/2) * m * r * [(T - mg) / m]
Simplifying further:
2T = r * (T - mg)
Finally, solving for T:
2T = Tr - mgr
T(2 - r) = mgr
T = mgr / (2 - r)
Substituting the known values, we have:
T = (0.2 kg)(9.8 m/s^2)(0.055 m) / (2 - 0.01 m)
Solving for T, we find the tension in the string.