A uniform bar 1m long is pivoted at the centre has a mass of 3kg. It is acted upon by a conquer which cause to possess an angular acceleration of 6rad/sec^2; calculate
a. The moment of inertia of the bar
b. The torque on the bar
A bar is 500 mm long and is stretched to 505 mm with a force of 50 KN. The bar is10mm diameter. Calculate the stress & strain.
To calculate the moment of inertia of the bar, we can use the formula:
Moment of Inertia (I) = (1/12) * m * L^2
where:
m = mass of the bar
L = length of the bar
In this case, the mass of the bar is given as 3 kg and the length is given as 1 m.
a. Calculating the moment of inertia:
I = (1/12) * 3 kg * (1 m)^2
I = 1/12 * 3 kg * 1 m^2
I = 1/4 kg·m^2
So, the moment of inertia of the bar is 1/4 kg·m^2.
b. To calculate the torque on the bar, we can use the formula:
Torque (τ) = I * α
where:
I = moment of inertia
α = angular acceleration
In this case, the moment of inertia (I) is 1/4 kg·m^2 and the angular acceleration (α) is given as 6 rad/sec^2.
b. Calculating the torque:
τ = (1/4 kg·m^2) * 6 rad/sec^2
τ = 6/4 kg·m^2·rad/sec^2
τ = 3/2 kg·m^2·rad/sec^2
So, the torque on the bar is 3/2 kg·m^2·rad/sec^2.