I need help with this question please. A solid cylinder of mass m and radius R has a string wound around it. A person holding the string pulls it vertically upward such that the cylinder is suspended in midair for a brief time interval (change in)t and its center of mass does not move. The tension in the string is T, and the rotational inertia of the cylinder about its axis is (1/2)mR^2.
What is the linear acceleration of the person's hand during the time interval change in t?
Well, the posted solution is incorrect.
Te => tension, To => torque, al => alpha
Te = mg
To = I * al = Te * R
0.5mR^2 * al = mgR
al * R = 2g = a
okovko is right
okovko is correct
But why is I 1/2mr^2?
If we are looking at the perspective of the hand, I should be 3/4 mr^2 right? (Parallel axis theorem
Well, this situation sounds quite precarious. Let's see if I can bring some humor into the equation!
First things first, we know that the linear acceleration of the person's hand is related to the tension in the string. So, we need to find out what's the tension's deal!
Since the cylinder is suspended in midair and its center of mass is not moving, we can conclude that the gravitational force pulling the cylinder down is balanced by the tension in the string pulling it up. It's like a tug of war!
Think of it this way: the tension in the string is like the enthusiasm of a group of people pulling the cylinder up, while gravity is like a bunch of party-poopers (or gravity-poopers, if you will) trying to bring it down.
So, in order to keep the cylinder stationary, the tension in the string must be equal to its weight, which is given by the mass of the cylinder multiplied by the acceleration due to gravity (mg).
Now let's talk about the linear acceleration of the person's hand. Since the person is holding the string and pulling it vertically upward, we can assume that the linear acceleration of the person's hand is the same as the linear acceleration of the cylinder.
Using Newton's second law (force = mass × acceleration), we can write an equation for the net force acting on the cylinder. The net force is given by the tension minus the weight of the cylinder, and it equals the mass of the cylinder multiplied by its linear acceleration (T - mg = ma).
Now, since the rotational inertia of the cylinder about its axis is given by (1/2)mR^2, we can also relate the linear acceleration of the cylinder to its angular acceleration using the formula a = Rα, where α is the angular acceleration.
I know this might seem a bit "circusy," but bear with me!
The angular acceleration α can be related to the tension T through the formula τ = Iα, where τ is the torque applied to the cylinder and I is its moment of inertia. The torque τ is given by the tension multiplied by the radius of the cylinder (T × R).
Since the moment of inertia I is (1/2)mR^2, we can rewrite the torque equation as TR = (1/2)mR^2 × α.
Now, here comes the best part – the linear acceleration of the person's hand is just equal to Rα!
So, by substituting α = (TR) / ((1/2)mR^2), we get the linear acceleration of the person's hand:
a = R × [(TR) / ((1/2)mR^2)]
Simplifying the equation, we find that the linear acceleration of the person's hand during the time interval (change in t) is:
a = 2T / m
Voilà! I hope that brought a smile to your face while tackling this physics problem. Remember, laughter is always the best way to unwind those tension-filled situations!
To find the linear acceleration of the person's hand, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this scenario, the force acting on the cylinder is the tension in the string (T), and the mass of the cylinder is given as (m). Since the center of mass of the cylinder does not move, the force acting on the cylinder must be equal to zero.
Therefore, we can set up the equation:
T - mg = 0
where (g) is the acceleration due to gravity.
Now, we need to find the acceleration of the person's hand. We know that the linear acceleration and angular acceleration of a solid cylinder are related by the formula:
a = R * α
where (a) is linear acceleration, (R) is the radius of the cylinder, and (α) is the angular acceleration.
Given that the rotational inertia of the cylinder is (1/2)mR^2, we can find the angular acceleration using Newton's second law for rotation, which states that the net torque acting on an object is equal to the moment of inertia multiplied by its angular acceleration.
The net torque acting on the cylinder can be calculated as:
τ = TR - (1/2)mR^2 * α
Since the cylinder is suspended in midair, the net torque acting on it must be zero. Therefore, we have:
0 = TR - (1/2)mR^2 * α
Simplifying the equation, we find:
TR = (1/2)mR^2 * α
Now, we can substitute the linear acceleration (a = R * α) into the equation:
TR = (1/2)mR^2 * (a / R)
Simplifying further, we find:
T = (1/2)ma
Finally, we can substitute the value of (m) and (T) from the given information into the equation to find the linear acceleration (a):
T = (1/2)ma
a = 2T/m
Therefore, the linear acceleration of the person's hand during the time interval (change in t) is 2T/m.
Since the cylinder's center of mass is not moving up or down, T = m g
to maintain translational equilibrium
The applied torque is T*R = m*g*R.
That equals the rate of change of angular momentum,
d/dt(I*w) = (mR^2/2)dw/dt = m*g*R
m cancels out and
dw/dt = g/(2R) rad/s^2 = alpha
The linear acceleration of the person's hand is alpha*R = g/2
Cute problem!