Question

Consider a bug which is crawling on the surface of a turntable rotating with constant angular velocity wo. List all the forces acting on it in a system of coordinates painted on the turntable (z is perpendicular to the turntable and the rotation is around the z-axis) and the direction of each of those forces, in each of the following cases: a) when the bug is moving out from the center of the turntable along a radius at constant speed v (with respect to the turntable) b) when the bug is moving on the top of the turnable at constant speed v (with respect to the turntable) in a circle of radius r concentric with the axis of rotation and in the direction of the rotation c) when the bug is moving on the top of the turnable at constant speed v in a circle of radius r concentric with the axis of rotation and opposite to the direction of the rotation. d) for the case a) imagine that at the instant the bug starts from the center at a constant speed of 4 cm/s somebody switches the speed of rotation of the turntable and that the lar velocity increases linearly from w- 33 rev/min to wy 45 rev/min, reaching the final value when the bug makes it to the edge of turntable (R 30 cm). Calculate the angu acceleration of the bug as a function of time. e) for cases b) and c) what value does the total force take when u takes the particular value wo r f) for the same two cases b) and c) assume that the coefficient of friction between the bug and the turntable is H, and find out the maximum speed with which the bug can move. u, and find out the maximum speed with which the b

Answer 1

In order to find the different components of forces, we will apply newton's second law of motion on the bug in the respective directions. From this by forming different equations from our available data we can get the acceleration.
Where N = normal force Fg = mg = Gravitational force F = mc = force due to motion mv = force as it going away from the centre F-s = static friction force on the centerfield force Applying Newton law N = mg 1 and (m w_0^2 - m_v) = mq 2 Where N = normal force mv = force due to motion mg = gravitational force F_s = central petal force F = ma Applying Newton�s law to the above system of the buy N = mg [which cancels each other) mw%2 R + m V Vector = ma 3 where R = the radius of the centre \r\n Where N = normal force mg = gravitational force f_s = centripetal force on static friction mv = force of motion Apply Newton�s law of motion N = mg and (m w_0^2 r - mv = ma) 4 from equation 2