Question

Can you please solve the problem with showing the details

MOHAMED ALDEGN 1996MB. Consider a thin uniform rod of mass M and length I, as shown above a. Show that the rotational inertia of the rod about an axis through its center and perpendicular to its length is MP/12. 8 vil: D.ALDEGNI ニレMLz The rod is now ghued to a thin hoop of mass M and radius R/2 to form a rigid assembly, as shoun above. The centers of the rod and the hoop concide at point P. T he assembly is mounted on a horizcotal axle through point P and perpendicular to the page b. What is the rotational inertia of the rod-hoop assembly about the axle? Several tuns of string are wrapped tightly around the circumference of the hoop. The system is at rest cat, also of mass M, grabs the free end of the string and hangs vertically from it without swinging as it unwinds, causing the rod-hoop assembly to rotate. Neglect friction and the mass of the string. c. Determine the tension T in the string d. Determine the angular acceleration a of the rod-boop assembly. MOHAMED ALDEGN e. Determine the linear acceleration of the cat. f After descending a distance H 53, the cat lets go of the string, At that instant, what is the angular momentum of the cat about point P?

Answer 1

Let be mass/ length = m/L mass of small element dx at distance x from the axis = M/L dx dI = M/L dx x^2 I = 2 integral M/L x^2 dx = 2 M/L (L/2)^3 times 1/3 = 1/12 ML^2 total I because of rod and because of ring I = ML^2/12 + M(R/2)^2 MR^2/12 + mR^2/4 = 1/3 MR^3 ma = Mg - T T R/2 = I alpha = MR^2/3 alpha T = 2/3 MR alpha alpha = a/R/2 = 2a/R T = 2/3 MR times 2a/R = $ma/3 m_a = Mg - 4Ma/3 = 7/3 ma = mg a = 3g/7 = > T = yM/3 times 3 g/7 = 4 mg/7 alpha a/R/2 = 2/R times 3g/6 = 6 g/7R a = 3g/7 velocity of cat the time of release = v^2 = u^2 + 2as v = squareroot 2 as = squareroot 2 times 3g/7 times 5l/7 = squareroot 10gl/7 L = mvr = mvR/2