# an object m = 3.05kg rolls fro

an object m = 3.05kg rolls from rest down an incline plane, itscenter initially at a height of 7.55m above the bottom of the ramp.The inclination of the incline as measured from the horizontal is37.3o. If the object is a solid uniform cylinder of radius r =1.25m, then at the bottom of the incline, what is the (a)rotational kinetic energy and (b) the angular momentum. If theobject is a hoop of radius, r = 1.05m, then at the bottom of theramp, what is the (c) rotational kinetic energy and (d) the angularmomentum?

Answer:

**a) Unifrom cyllinder**

**Let w is the angular speed and v is the linear speed atthe bottom.**

**Apply conservation of energy**

**(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h**

**(1/2)*m*v^2 + (1/2)*(1/2)*m*r^2*w^2 = m*g*h**

**(1/2)*m*v^2 + (1/4)*m*(r*w)^2 = m*g*h**

**(1/2)*m*v^2 + (1/4)*m*v^2 = m*g*h**

**(3/4)*m*v^2 = m*g*h**

**v = sqrt(4*g*h/3)**

**w = v/r**

**= sqrt(4*g*h/3)/r**

**= sqrt(4*9.8*7.55/3)/1.25**

**= 7.94 rad/s**

**KE_rotationa = (1/2)*I*w^2**

**= (1/2)*(1/2)*m*r^2*w^2**

**= (1/4)*3.05*1.25^2*7.94^2**

**= 75.1 J**

**b) angular momentum = I*w**

**= (1/2)*m*r^2*w**

**= (1/2)*3.05*1.25^2*7.94**

**= 18.9 kg.m^2/s**

**c) Hoop**

**Let w is the angular speed and v is the linear speed atthe bottom.**

**Apply conservation of energy**

**(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h**

**(1/2)*m*v^2 + (1/2)*m*r^2*w^2 = m*g*h**

**(1/2)*m*v^2 + (1/2)*m*(r*w)^2 = m*g*h**

**(1/2)*m*v^2 + (1/2)*m*v^2 = m*g*h**

**m*v^2 = m*g*h**

**v = sqrt(g*h)**

**w = v/r**

**= sqrt(g*h)/r**

**= sqrt(9.8*7.55)/1.05**

**= 8.19 rad/s**

**KE_rotationa = (1/2)*I*w^2**

**= (1/2)*m*r^2*w^2**

**= (1/2)*3.05*1.05^2*8.19^2**

**= 113 J**

**d) angular momentum = I*w**

**= m*r^2*w**

**= 3.05*1.05^2*8.19**

**= 27.5 kg.m^2/s**