Question

Rank the following cases in terms of lowest torque to highest torque: - a. b. c. d.

A 5 N force applied to a 1.0 m lever arm at 30 degrees.
- a. b. c. d.
A 0.5 kg-m2 disk accelerating at 10.0 rad/s2
- a. b. c. d.
A 10 N force applied to a 0.6 m lever arm at 90 degrees.
- a. b. c. d.
A 1.0 kg-m2 disk accelerating at 1.0 rad/s2

Answer 1

Answer:

$T_C>T_B>T_A>T_D$

Explanation:

Now we calculate the torque in each case:

A.

Given:

Force applied to the lever arm, $F=5\ N$

distance from the pivot, $r=1\ m$

angle of application, $\theta=30^{\circ}$

Torque:

$T_A=5\times 1\times sin\ 30^{\circ}$

$T_A=2.5\ N.m$

B.

Given:

mass moment of inertia of a disc, $I=0.5\ kg.m^2$

acceleration of the disc, $\alpha=10\ rad.s^{-2}$

Torque:

$T_B=I.\alpha$

$T_B=0.5\times 10$

$T_B=5\ N.m$

C.

force applied to the lever, $F=10\ N$

distance of lever arm, $r=0.6\ m$

angle of application of force, $\theta = 90^{\circ}$

Torque:

$T_C=F.r.sin\ \theta$

$T_C=10\times 0.6\times sin\ 90^{\circ}$

$T_C=6\ N.m$

D.

mass moment of inertia of a disc, $I=1\ kg.m^2$

acceleration of the disc, $\alpha=1\ rad.s^{-2}$

Torque:

$T_D=I.\alpha$

$T_D=1\times 1$

$T_D=1\ N.m$