A clock has a second hand whose tip rubs against the inside of the glass cover. If the frictional force between the glass cover and the tip of the second hand is 0.0010 N and the length of the hand is 6.5 cm, what is the minimum torque that must be applied to the second hand if the clock is not to be stopped?
.001 * .065 Nm
To find the minimum torque required to prevent the clock from stopping, we need to consider the balance of torques on the second hand.
Torque, denoted as τ, is the rotational equivalent of force and is calculated as the product of the force applied and the distance from the axis of rotation. In this case, the axis of rotation is where the second hand is attached to the clock mechanism, and the force is the frictional force between the glass cover and the tip of the second hand.
The formula for torque is given by:
τ = F * r
Where:
τ = torque
F = force
r = distance from the axis of rotation
In this question, we are given the frictional force between the glass cover and the tip of the second hand as 0.0010 N. The length of the hand is 6.5 cm, which is equivalent to 0.065 meters.
Substituting the values into the formula, we have:
τ = 0.0010 N * 0.065 m
Calculating this, we find:
τ = 0.000065 Nm
Therefore, the minimum torque that must be applied to the second hand to prevent the clock from stopping is 0.000065 Nm.