A yo-yo of radius 10 cm and mass 1 kg is dropped from rest. What is the instantaneous torque on the yo-yo if I hold the string with my fingers?
To find the instantaneous torque on the yo-yo, we need to consider the forces acting on it. When the yo-yo is dropped from rest, the only force acting on it is gravity.
The torque on an object is given by the product of the force applied and the perpendicular distance from the point of application to the axis of rotation.
In this case, as you hold the string with your fingers, you provide a force in the upward direction to counteract the weight of the yo-yo. This force is equal in magnitude and opposite in direction to the weight of the yo-yo, which is the force due to gravity acting downward.
The torque you apply to the yo-yo is given by the equation:
Torque = Force x Distance from axis of rotation
The distance from the axis of rotation to the point where you hold the string is the radius of the yo-yo (10 cm or 0.1 m).
The force you apply is equal to the weight of the yo-yo, which can be calculated using the equation:
Force = Mass x Acceleration due to gravity
Given that the mass of the yo-yo is 1 kg and the acceleration due to gravity is approximately 9.8 m/s^2, the force is:
Force = 1 kg x 9.8 m/s^2 = 9.8 N
Finally, we can calculate the instantaneous torque on the yo-yo:
Torque = Force x Distance from axis of rotation
= 9.8 N x 0.1 m
= 0.98 Nm
Therefore, the instantaneous torque on the yo-yo is 0.98 Nm.