At 15 minutes past the hour, a 47.0 g mouse runs up a grandfather clock and sits himself down at the end of the minute-hand. If the minute-hand for the clock is 16.0 cm long, what is the magnitude of the torque exerted by the mouse on the minute-hand? Give your answer in the standard units of N m.
mass = 0.047 kg
weight = 0.047 * 9.81 = 0.461 Newtons
lever arm = 0.16 meter
toque = 0.461 * 0.16 Nm
To determine the magnitude of the torque exerted by the mouse on the minute-hand, we need to use the formula:
Torque = Force × Distance × sin(θ),
where "Force" is the force applied by the mouse, "Distance" is the distance from the point of rotation to the application of force, and "θ" is the angle between the force vector and the lever arm.
In this case, the force applied by the mouse is the weight of the mouse, given by:
Force = mass × gravity,
where "mass" is the mass of the mouse and "gravity" is the acceleration due to gravity.
Using the given values:
mass = 47.0 g = 47.0 × 10^(-3) kg (converting grams to kilograms),
gravity = 9.8 m/s^2 (standard value).
Force = 47.0 × 10^(-3) kg × 9.8 m/s^2 = 0.4616 N (to four decimal places).
Next, we need to find the lever arm, which is the distance from the point of rotation (center of the clock face) to the application of force (the mouse). The lever arm is the radius of the circle traced by the end of the minute-hand, which is given as 16.0 cm.
Distance = 16.0 cm = 16.0 × 10^(-2) m (converting centimeters to meters) = 0.16 m.
Now, we can calculate the magnitude of the torque:
Torque = 0.4616 N × 0.16 m × sin(θ).
However, we need to determine the angle θ between the force vector and the lever arm. In this case, the mouse is sitting on the minute-hand, which is perpendicular to the clock face. Therefore, θ is 90 degrees (π/2 radians).
Torque = 0.4616 N × 0.16 m × sin(π/2) = 0.4616 N × 0.16 m × 1 = 0.07385 N·m (to five decimal places).
Therefore, the magnitude of the torque exerted by the mouse on the minute-hand is approximately 0.07385 N·m.