Question

A uniform rod is set up so that it can rotate about an axis at perpendicular to one of its ends. The length and mass of the rod are 0.759 m and 1.03 kg , respectively. A force of constant magnitude ???? acts on the rod at the end opposite the rotation axis. The direction of the force is perpendicular to both the rod's length and the rotation axis. Calculate the value of ???? that will accelerate the rod from rest to an angular speed of 6.87 rad/s in 9.95 s

Answer 1

Answer:

F = 0.179 N

Explanation:

given,

length of the rod = 0.759 m

mass of the rod = 1.03 kg

force = ?

initial angular speed = 0 rad/s

final angular speed = 6.87 rad/s

time interval = 9.95 s

the angular acceleration

$\alpha = \dfrac{\omega_f-\omega_i}{\Delta t}$

$\alpha = \dfrac{6.87 -0}{9.95}$

$\alpha =0.690\ rad/s^2$

moment inertial of the rod

$I =\dfrac{mL^2}{3}$

torque produced by the rod τ = I α

torque is also equal to τ = F L

now,

F L =  I α

$F L=(\dfrac{mL^2}{3})\times \alpha$

$F= (\dfrac{1.03\times 0.759}{3})\times 0.690$

F = 0.179 N

Answer 2

Answer:

Force will be 0.179 N

Explanation:

We have given length of the rod l = 0.759 m

Mass m = 1.03 kg

We have to find the constant force

Initial angular velocity $\omega _i=0rad/sec$

And final angular velocity $\omega _f=6.87rad/sec$

Time t = 9.95 sec

So angular acceleration $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{6.87-0}{9.95}=0.69rad/sec^2$

Moment of inertia of rod $I=\frac{ml^2}{3}=\frac{1.03\times 0.759^2}{3}=0.1977kgm^2$

So torque $\tau =I\alpha =0.1977\times 0.69=0.1364N-m$

We know that torque $\tau =Fl$

So $0.1364=F\times 0.759$

F = 0.179 N