Spring Constant - General Physics - Solved Past Paper, Exams of Physics

Download Spring Constant - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 6. (25 pts) A spring of spring constant k is attached to a point on the outer rim of a solid disc (moment of inertia I = (1/2)MR2) with mass m and radius R. The disc can spin freely about an axis through its center. As the disc rotates, the spring stretches, as shown below. R k R k θ Unstretched Stretched a) (5 pts) By what distance does the point at which the spring is attached to the disc move when the disc rotates by an angle θ? s = Rθ b) (5 pts) After rotating by angle θ, how much force does the spring apply to the disc (written in terms of the angle)? The spring is now stretched by the amount listed in part (a), so F = −kx = −kRθ c) (5 pts) What is the torque applied to the disc by the spring at this point? τ = Fr sin φ = (−kRθ)(R) = −kR2θ Note that the force exerted by the spring at the point of attachment is tangent to the edge of the disc, thus φ = 90◦. d) (5 pts) Using “Newton’s 2nd Law for Rotation,” write down the (differential) equation of motion for this system (but do not solve it). Write it so that all variables relating to the rotation of the disc (angular position, angular velocity, and angular acceleration) are in terms of θ. (Hint: First write down Newton’s 2nd Law for Rotation in general, then substitute for each of the pieces from your answers above.) τ = Iα −kR2θ = 1 2 MR2 d2θ dt2 d2θ dt2 + 2k M θ = 0 e) (5 pts) From your equation of motion, what should be the angular frequency of the motion if the disc is now released? The value of ω was taken from the coefficient of the x or θ in previous examples. The coefficient in this case is 2k/M . Thus ω = √ 2k M