Answer:
a) To calculate the torque from the engine to achieve an angular acceleration of 30 rad/s², we use the formula:
\[ \text{Torque} = \text{Inertia} \times \text{Angular Acceleration} \]
Given:
Inertia of engine (Ie) = 20 kg m²
Angular Acceleration (α) = 30 rad/s²
\[ \text{Torque} = 20 \times 30 \]
\[ \text{Torque} = 600 \text{ Nm} \]
b) To calculate the torque needed from the engine to achieve an angular acceleration of 30 rad/s² considering the frictional torque, we use the formula:
\[ \text{Torque} = \text{Inertia} \times \text{Angular Acceleration} + \text{Frictional Torque} \]
Given:
Inertia of drive shaft (Id) = 30 kg m²
Frictional Torque = 200 Nm
\[ \text{Torque} = (20 + 860) \times 30 + 200 \]
\[ \text{Torque} = 26400 + 200 \]
\[ \text{Torque} = 26600 \text{ Nm} \]
c) The total kinetic energy of the engine and load three seconds after starting from rest can be calculated using the formula for kinetic energy:
\[ KE = \frac{1}{2} I \omega^2 \]
Where:
- I is the inertia
- ω is the angular velocity
Given:
Inertia of engine (Ie) = 20 kg m²
Inertia of load (Il) = 860 kg m²
Angular acceleration (α) = 30 rad/s²
Time (t) = 3 seconds
The final angular velocity (ω) can be calculated using the equation:
\[ ω = αt \]
\[ ω = 30 \times 3 = 90 \text{ rad/s} \]
Now, we can calculate the kinetic energy for both the engine and the load:
\[ KE_{engine} = \frac{1}{2} \times 20 \times (90)^2 \]
\[ KE_{engine} ≈ 81000 \text{ J} \]
\[ KE_{load} = \frac{1}{2} \times 860 \times (90)^2 \]
\[ KE_{load} ≈ 3483000 \text{ J} \]
The total kinetic energy is the sum of the kinetic energies of the engine and the load:
\[ KE_{total} = KE_{engine} + KE_{load} \]
\[ KE_{total} ≈ 81000 + 3483000 \]
\[ KE_{total} ≈ 3564000 \text{ J} \]
d) The instantaneous power expended by the engine can be calculated using the formula:
\[ P = \text{Torque} \times \text{Angular Velocity} \]
Given:
Torque = 26600 Nm (from part b)
Angular velocity (ω) = 90 rad/s (from part c)
\[ P_{engine} = 26600 \times 90 \]
\[ P_{engine} ≈ 2394000 \text{ W} \]
The frictional power dissipated can be calculated using the frictional torque:
\[ P_{friction} = \text{Frictional Torque} \times \text{Angular Velocity} \]
Given:
Frictional Torque = 200 Nm
\[ P_{friction} = 200 \times 90 \]
\[ P_{friction} = 18000 \text{ W} \]
e) If the load consists additionally of a mass being lifted from the ground, the kinetic energy of the engine system would be increased because lifting the additional mass requires additional work, which increases the total kinetic energy of the system.
