A 12N force is applied to a 20kg door If a frictional torque is present what will be the angular acceleration of the door The door is 80cm wide and I=1/3ml^2.
torque= I*angacc
12*.8=1/3 m (.8^2)*angacceleration
A 12N force is applied to a 20kg door If a 5m.N frictional torque is present what will be the angular acceleration of the door The door is 80cm wide and I=1/3ml^2.
you are correct, I did not subtract frictional torque.
12*.8-.5= .....
To find the angular acceleration of the door, we need to apply Newton's second law for rotational motion. The formula is given by:
τ = I * α
Where:
τ is the torque applied to the object,
I is the moment of inertia of the object,
α is the angular acceleration of the object.
In this case, we know the force applied to the door (12N) and the width of the door (80cm). To find the torque, we can use the formula:
τ = F * r
Where:
F is the force applied,
r is the length of the lever arm.
In this case, the lever arm is half of the door's width since the force is applied at the center of the door. Therefore, the lever arm (r) is 80cm / 2 = 40cm = 0.4m.
Now, we can calculate the torque:
τ = F * r
τ = 12N * 0.4m
τ = 4.8 Nm
Next, we need to find the moment of inertia (I) of the door. The moment of inertia depends on the shape and mass distribution of the object. In this case, the moment of inertia can be calculated using the formula for a rectangular door:
I = (1/3) * m * (l^2)
Where:
m is the mass of the door (20kg),
l is half the width of the door. In this case, l = 80cm / 2 = 40cm = 0.4m.
Now, we can calculate the moment of inertia:
I = (1/3) * m * (l^2)
I = (1/3) * 20kg * (0.4m)^2
I = 1.0667 kg·m²
Finally, we can find the angular acceleration by rearranging the torque formula:
τ = I * α
α = τ / I
α = 4.8 Nm / 1.0667 kg·m²
Calculating this will give us the angular acceleration of the door.