starting from rest s=0,pulley A (rA=50 mm)is given a constant angular acceleration, A=6 rad/s2. Pulley C(rC=150mm) has an inner hub D (rD=75 mm) which is fixed to C and turns with it. What is the speed of block B when it has risen s=6 m ?

where is the answer

To find the speed of block B when it has risen s = 6 m, we need to calculate the angular velocity of pulley A and then use it to find the linear velocity of block B.

Step 1: Find the initial angular velocity of pulley A.
Given: Angular acceleration of pulley A, A = 6 rad/s^2
Using the formula: ω = ω0 + At
Since the initial angular velocity, ω0, is 0 rad/s (starting from rest), we can simplify the equation to:
ω = At

Substituting the values:
ω = 6 rad/s^2 * t

Step 2: Find the time taken by block B to rise 6 m.
The distance traveled, s = 6 m
Using the formula: s = ut + 0.5At^2
Since the initial velocity, u, is 0 m/s (starting from rest), we can simplify the equation to:
s = 0.5At^2

Rearranging the equation:
t^2 = 2s / A

Substituting the values:
t^2 = 2 * 6 m / 6 rad/s^2
t^2 = 2 seconds^2
t = √(2) seconds

Step 3: Calculate the final angular velocity of pulley A.
Using the equation:
ω = At

Substituting the values:
ω = 6 rad/s^2 * √(2) seconds
ω = 6√(2) rad/s

Step 4: Calculate the linear velocity of block B.
The linear velocity of block B is equal to the linear velocity of pulley C since they are connected.
Using the formula: v = ω * r
v = 6√(2) rad/s * 0.15 m
v = 0.9√(2) m/s

Therefore, the speed of block B when it has risen s = 6 m is 0.9√(2) m/s.

To find the speed of block B when it has risen s = 6 m, we need to determine the angular displacement of pulley A, the angular displacement of pulley C, and the linear displacement of block B. From these values, we can calculate the speed of block B.

First, let's calculate the angular displacement of pulley A. We know that it starts from rest and has a constant angular acceleration, A = 6 rad/s^2. We can use the following kinematic equation:

θ = ω0*t + (1/2)*A*t^2

Since pulley A starts from rest, ω0 (initial angular velocity) is 0. Plugging in the values, we get:

θ = (1/2)*A*t^2

Since the angular displacement of pulley A is equal to the linear displacement of block B (s = θ*rA), we can substitute s = θ*rA:

s = (1/2)*A*t^2*rA

Now, we need to find the time it takes for the block to rise s = 6 m. We can use the following kinematic equation:

s = ω0*t + (1/2)*A*t^2

Since the initial angular velocity ω0 is 0, the equation simplifies to:

s = (1/2)*A*t^2

Plugging in the known values:

6 = (1/2)*6*t^2

Simplifying the equation:

12 = 6*t^2

Dividing both sides by 6:

2 = t^2

Taking the square root:

t = √2 seconds

Now, we can substitute the value of t = √2 seconds into the equation for s = (1/2)*A*t^2*rA:

s = (1/2)*6*(√2)^2*50

s = 150 mm = 0.15 m

This means that when block B has risen 6 m, the angular displacement of pulley A is equivalent to a linear displacement of 0.15 m.

Next, let's calculate the angular displacement of pulley C. Pulley C is connected to pulley A through the fixed inner hub D. Since D has a radius of 75 mm, which is half of the radius of pulley C, the angular displacement of pulley C is double that of pulley A:

θC = 2*θA

Using the value of the angular displacement from pulley A, we can calculate:

θC = 2*0.15 m / rC

θC = 0.15 / 0.15 = 1 rad

Now, let's calculate the linear displacement of block B. The linear displacement of block B can be determined by the circumference of pulley C, multiplied by the number of rotations pulley C has made:

sB = 2π*rC*nC

The number of rotations nC can be calculated as the angular displacement of pulley C divided by 2π:

nC = θC / 2π

Plugging in the known value of θC, we get:

nC = 1 rad / (2π)

nC ≈ 0.159 rotations

Now, we can calculate the linear displacement of block B:

sB = 2π * 0.15 m * 0.159

sB ≈ 0.15 m

Therefore, when block B has risen s = 6 m, the linear displacement of block B is approximately 0.15 m.

Finally, we can calculate the speed of block B. The speed v is defined as the rate of change of the linear displacement:

v = Δs / Δt

Since the linear displacement Δs is 0.15 m and the time Δt is √2 seconds, we can substitute these values:

v = 0.15 m / √2 s

Calculating this value, we get:

v ≈ 0.106 m/s

Therefore, the speed of block B when it has risen s = 6 m is approximately 0.106 m/s.