A large iron cylinder of mass 380 kg rests so that its circular face rolls along the edge of table as shown in the figu…

Question

(a) calculate the moment of inertia and the total mass of the entire assembly.
moment of inertia = ? kg.m2
total mass = ? kg

(b) What is the total kinetic energy of the system after the attached mass descends through a height h ? = ? J

(c) If the entire assembly rolls without slipping, what is its velocity at this time ? cm/s

(d) How far has the system moved along the horizontal ? cm
http://d.yimg.com/hd/answers/i/5ccc9a3d2bb74e50a71cb9ceb1d8a964_A.png?a=answers&mr=0&x=1382388308&s=56deec68f0bc3e4accf4c42ef9b4ce45
(a) calculate the moment of inertia and the total mass of the entire assembly.
moment of inertia = ? kg.m2
total mass = ? kg

(b) What is the total kinetic energy of the system after the attached mass descends through a height h ? = ? J

(c) If the entire assembly rolls without slipping, what is its velocity at this time ? cm/s

(d) How far has the system moved along the horizontal ? cm
http://d.yimg.com/hd/answers/i/5ccc9a3d2bb74e50a71cb9ceb1d8a964_A.png?a=answers&mr=0&x=1382388308&s=56deec68f0bc3e4accf4c42ef9b4ce45 calculate the moment of inertia and the total mass of the entire assembly.moment of inertia = ? kg.m2 total mass = ? kg What is the total kinetic energy of the system after the attached mass descends through a height h ? = ? If the entire assembly rolls without slipping, what is its velocity at this time ? cm/s How far has the system moved along the horizontal ? cm http://d.yimg.com/hd/answers/i/5ccc9a3d2bb74e50a71cb9ceb1d8a964_A.png?a=answers&mr;=0&x;=1382388308&s;=56deec68f0bc3e4accf4c42ef9b4ce45

Explanation / Answer

I = 1/2 M R^2 + M R^2 + 1/2 m r^2 + m R^2

Moment of inertia about point of contact with table

using the parallel axis theorem.

Torque = T r = I a     where a is the "angular acceleration"

We're using the angular acceleration about the point of contact

which is the same as the angular acceleration about the center of mass

m2 g - T = m2 a r     where T is tension in cable and m2 the hanging mass

Eliminate T (add equations) to get a (angular acceleration)

KE = 1/2 I w^2 + 1/2 m2 (r * w)2      since r * w = v

h = 1/2 (r * a) t^2 to get time     where linear acceleration = r * a

w = a t    where w is the angular velocity

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