A small block with mass m slides along a frictionlesshorizontal surface with a speed v =with and sticks to the end of a uniform rod (mass M = m,length L = 0.75 m) that can rotate around a frictionlessaxle through its upper end as shown.:0.4 m/s. It collides%3Da) Use Newton's 2nd law to find the angular frequency wof small angle oscillations for the combined system.b) What is the amplitude (maximum displacement 0max)from the vertical, expressed in degrees? Lm v

The mass m is moving with a velocity v towards the rod, and the rod is initially at rest After some time, the mass sticks to the rod, and since the rod can rotate about an axle, the system of rod and mass will gain some angular velocity, with a tangential velocity equal to (m1+m2)v=m1v1+m2v2This comes from the law of conservation of momentumsince the rod was initially at rest, v2=0And, m1=m2=m2mv=mv1v=12v1v=120.4v=0.2 m/s So, the tangential velocity of the system is equal to 0.2 m/sThe angular velocity frequency ω can be found using Newtons second law of rotationτ=Iατ is the torque acting on the systemI is the moment of inertia of the systemα is the angular accelerationTorque is given by the product of the force F acting on the object and the perpendicular distance from the axis to the forceτ=F×Lτ=-2mgsinθL2mgsinθ is the component of force acting on the total system mass of 2m and perpendicular to the length LNegative sign indicates that the applied force is in the opposite direction of the torque The moment of inertia for the system of mass 2m isI=2mL2And the angular accelration isα=d2θdt2So,-2mgLsinθ=2mL2d2θdt2On re-arranging the terms, we get a second order differential equation in θLd2θdt2+gsinθ=0d2θdt2+gLsinθ=0For small angle of oscillations, sinθ≈θd2θdt2+gLθ=0The solution of the above equation is obtained in the form of a cos function asθ=θosin(gLt)The constant gL is the angular frequency of oscillationsSo, ω=gLω=9.80.75ω=3.615 rad/s This is the angular frequency of oscillations as computed by Newton's second lawThe displacement θ is given byθ=θosin(ωt)This displacement θ will be maximum when sin(ωt)=1This will happen when ωt=π2t=π2ωt=3.142×3.615t=0.4343 s This is the time required to achieve maximum displacement For a tangential velocity 0.2 m/s, the distance covered in time 0.4343 s isDistance=0.2×0.4343Distance=0.087 mThis is the distance covered by the arc during oscillationsNow, the angular displacement corresponding to this distance on the arc isθ=0.087Lθ=0.0870.75θ=0.1158 radθ=0.1157×180π°θ=6.64° So, the maximum angular displacement from the vertical is 6.64 degreesAnswers: The angular frequency of oscillations of the combined system is 3.615 rad/s The maximum angular displacement from the vertical is 6.64o

Answered: A small block with mass m slides along…