A block (mass = 2.9 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1.2 x 10-3 kg·m2), as the figure shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of 0.033 m during the block's descent. Find (a) the angular acceleration of the pulley and (b) the tension in the cord
A: Well, since you mentioned a figure, I must say that the angular acceleration of the pulley is quite artistic. It twirls and swirls, just like a ballerina on stage. Simply breathtaking!
B: As for the tension in the cord, it's feeling quite stressed these days. I mean, it has to support the weight of the block and still keep the pulley spinning. Talk about multitasking! But don't worry, with a little relaxation and some positive affirmations, I'm sure the tension will be just fine.
To find the angular acceleration of the pulley, we can use the principle of conservation of energy.
Step 1: Find the gravitational potential energy of the block before it starts falling.
GPE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height the block is above the ground.
GPE = (2.9 kg)(9.8 m/s^2)(0.033 m)
Step 2: Find the change in the gravitational potential energy as the block falls.
We know that the block falls a distance equal to the radius of the pulley (0.033 m).
So, the change in height is h = -0.033 m (negative because the block is falling).
Change in GPE = mgh = (2.9 kg)(9.8 m/s^2)(-0.033 m)
Step 3: Find the change in kinetic energy of the block.
Kinetic energy is given by KE = 0.5mv^2, where m is the mass of the block and v is its velocity.
Since the block is initially stationary and starts from rest, its initial velocity is 0 m/s.
Thus, the change in kinetic energy is equal to the final kinetic energy of the block.
Change in KE = 0.5mv^2 = 0.5(2.9 kg)(0)^2
Step 4: Apply the principle of conservation of energy.
Change in GPE + Change in KE = 0
(2.9 kg)(9.8 m/s^2)(-0.033 m) + 0.5(2.9 kg)(0)^2 = 0
Step 5: Solve for the angular acceleration of the pulley.
Since the moment of inertia of the pulley is given (1.2 × 10^-3 kg·m^2), we can use the equation:
Change in GPE + Change in KE = (1/2)Iω^2, where ω is the angular velocity of the pulley.
(2.9 kg)(9.8 m/s^2)(-0.033 m) = (1/2)(1.2 × 10^-3 kg·m^2)ω^2
Step 6: Solve for the angular acceleration.
The angular acceleration (α) is related to angular velocity (ω) by the equation: α = ω/t, where t is the time taken for the block to fall the distance.
Rearranging the equation in step 5, we have:
(2.9 kg)(9.8 m/s^2)(-0.033 m) = (1/2)(1.2 × 10^-3 kg·m^2)ω^2 / t
Simplifying and rearranging, we get:
α = (2.9 kg)(9.8 m/s^2)(-0.033 m) / [(1/2)(1.2 × 10^-3 kg·m^2)(t)]
To find the tension in the cord, we can use the equation for torque.
Step 7: Find the torque applied to the pulley.
Torque is given by the equation: τ = Iα, where I is the moment of inertia of the pulley and α is the angular acceleration.
τ = (1.2 × 10^-3 kg·m^2) * α (using the value of α from step 6)
Step 8: Find the tension in the cord.
We can use the equation for torque: τ = Fr, where F is the tension in the cord and r is the radius of the cord around the pulley.
τ = Fr = (2.9 kg)(9.8 m/s^2)(0.033 m)
Solving for F, we get:
F = (2.9 kg)(9.8 m/s^2)(0.033 m) / r
So, the step-by-step solution is as follows:
(a) The angular acceleration of the pulley is given by:
α = (2.9 kg)(9.8 m/s^2)(-0.033 m) / [(1/2)(1.2 × 10^-3 kg·m^2)(t)]
(b) The tension in the cord is given by:
F = (2.9 kg)(9.8 m/s^2)(0.033 m) / r
To find the angular acceleration of the pulley, we can use the equation:
α = (2τ)/(mR^2 + I)
Where:
- α is the angular acceleration of the pulley,
- τ is the torque acting on the pulley,
- m is the mass of the block,
- R is the radius of the cord around the pulley, and
- I is the moment of inertia of the pulley.
To find the torque, we need to first calculate the force causing it. In this case, the force is the weight of the block, given by:
F = mg
Where:
- F is the force,
- m is the mass of the block, and
- g is the acceleration due to gravity.
Now, we can calculate the torque by multiplying the force by the radius of the pulley:
τ = rF
Where:
- τ is the torque,
- r is the radius of the pulley, and
- F is the force.
Finally, substituting the values into the equation for angular acceleration:
α = (2τ)/(mR^2 + I)
We can now solve for α.
To find the tension in the cord, we can use the relationship between tension, mass, and acceleration. The tension in the cord is equal to the force applied to the block, which is given by its weight. So:
T = mg
Where:
- T is the tension in the cord,
- m is the mass of the block, and
- g is the acceleration due to gravity.
Now, substituting the given values into the equation for tension:
T = mg
We can calculate the tension in the cord.