Answer:
a) τ = 0.672 N m , b) θ = 150 rad , c) W = 100.8 J
Explanation:
a) for this part let's start by finding angular acceleration, when the angular velocity stops it is zero (w = 0)
w = w₀ + α t
α = -w₀ / t
α = 120 / 2.5
α = 48 rad / s²
The moment of inertia of a cylinder is
I = ½ M R²
Let's calculate the torque
τ = I α
τ = ½ M R² α
τ = ½ 2.8 0.1² 48
τ = 0.672 N m
b) we look for the angle by kinematics
θ = w₀ t + ½ α t2
θ = ½ α t²
θ = ½ 48 2.5²
θ = 150 rad
c) work in angular movement
W = τ θ
W = 0.672 150
W = 100.8 J
The constant torque that will bring this grinding wheel from rest is 0.672 Newton.
Given the following data:
Mass = 2.8 kg.
Radius = 0.1 m.
Initial angular velocity = 0 rad/s (since it's starting from rest).
Final angular velocity = 120 rad/s.
Time = 2.5 seconds.
First of all, we would determine the angular acceleration of this grinding wheel by using this formula:
[tex]\alpha =\frac{\omega_f - \omega_i}{t} \\\\\alpha =\frac{120 - 0}{2.5}\\\\\alpha =48\;rad/s^2[/tex]
For the torque:
[tex]\tau = I\alpha \\\\\tau =\frac{1}{2} mr^2\alpha \\\\\tau =\frac{1}{2} \times 2.8 \times 0.1^2 \times 48[/tex]
Torque = 0.672 Newton.
Mathematically, the angle is given by the second equation of kinematics:
[tex]\theta = \omega_it + \frac{1}{2} \alpha t^2\\\\\theta = 0(2.5) + \frac{1}{2} \times (48)\times 2.5^2\\\\\theta =24 \times 6.25[/tex]
Angle = 150 rad.
Mathematically, the work done by the torque is given by this formula:
[tex]W = \tau \theta\\\\W= 0.672 \times 150[/tex]
W = 100.8 Joules.
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