The rotor of an electric motor has rotational inertia im = 7.17 x 10-3 kg·m2 about its central axis. the motor is used to change the orientation of the space probe in which it is mounted. the motor axis is mounted along the central axis of the probe; the probe has rotational inertia ip = 6.84 kg·m2 about this axis. calculate the number of revolutions of the rotor required to turn the probe through 44.1° about its central axis.

the number of revolutions of the rotor required to turn the probe through 44.1° about its central axis is approximately 117 revolution.

Step-by-step explanation:

From the conservation of angular momentum of the motor

Ip× theta p = Im × theta m

Where IP = 6.84 is the rotational inertia of the probe along it central axis

Theta p= 44.1°, Im = 7.17 ×10^-3 is the rotational inertia of the electric motor about it central axis.

Ip× theta p = Im × theta m

6.84 × 44.1 = 0.00717 × theta m

301.644 = 0.00717 × theta m

Theta m= 301.644 /0.00717

= 42,070.3°

Number of revolution =42,070.3°/360

= 117 revolution

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.