There is an analogy between quantities in rotational and translational dynamics - AQA - A-Level Physics - Question 1 - 2017 - Paper 6

To find the total moment of inertia about the axis of rotation, we sum the moment of inertia of the jib and the contributions from the trolley and load:

1. Moment of inertia of the jib: $I_{jib} = 2.6 imes 10^7 ext{ kg m}^2$

2. Moment of inertia of the trolley: $I_{trolley} = m_{trolley} imes r^2 = (2.2 imes 10^3 ext{ kg}) imes (35 ext{ m})^2 = 2.2 imes 10^3 imes 1225 ext{ kg m}^2$
   $I_{trolley} = 2.695 imes 10^6 ext{ kg m}^2$

3. Moment of inertia of the load: Assuming the load has the same mass as the trolley, we calculate the same way: $I_{load} = m_{load} imes r^2 = (2.2 imes 10^3 ext{ kg}) imes (35 ext{ m})^2 = 2.695 imes 10^6 ext{ kg m}^2$

4. Total moment of inertia: $I_{total} = I_{jib} + I_{trolley} + I_{load} = 2.6 imes 10^7 + 2.695 imes 10^6 + 2.695 imes 10^6$ $I_{total} = 2.9 imes 10^7 ext{ kg m}^2$

To calculate the maximum angular speed of the jib, we use the area under the angular speed versus time graph:

1. The total area under the graph represents the angular displacement, which can be represented as: ext{Angular displacement} = rac{1}{2} imes ext{height} imes ext{base}

2. The area consists of two triangles:

   • First triangle (0 to 50 s): ext{Area}_{1} = rac{1}{2} imes 50 ext{ s} imes ext{max height}
   • Second triangle (50 to 95 s): ext{Area}_{2} = rac{1}{2} imes (95-50) ext{ s} imes ext{max height}

3. Thus, the total area giving angular displacement is: heta = (rac{1}{2} imes 50 imes ext{ω}_{max}) + (rac{1}{2} imes 45 imes ext{ω}_{max}) = 4.7 ext{ rad}

4. Solving for ω_max: ext{ω}_{max} = rac{2 imes 4.7}{50 + 45} = rac{9.4}{95} ext{ rad/s} ext{ yielding } 0.099 ext{ rad/s}

When the motor is disengaged and a constant braking torque is applied, the total moment of inertia increases due to the trolley moving outward.

Using the relationship between angular deceleration and torque, let:

• Torque = moment of inertia × angular deceleration

Given that the braking torque remains constant, the time taken can be determined:

• Assume it takes $T$ seconds to stop:

  • The equation becomes: $ext{ω}_{max} - ext{ω}_{final} = - ext{angular deceleration} imes T$

• Since the final angular speed is zero, we find: T = rac{ ext{ω}_{max}}{ ext{angular deceleration}}

Using the values calculated previously, you can substitute to find the exact stopping time for the jib.