Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 weeks ago

# Physics Problem HW#8?

A hollow sphere is attached to the end of a uniform rod. The sphere has a radius of

0.68

m and a mass of

0.48

kg. The rod has a length of

1.22

m and a mass of

0.56

kg. The rod is placed on a fulcrum (pivot) at

X

=

0.44

m from the left end of the rod.

(a) Calculate the moment of inertia (click for graphical table) of the contraption around the fulcrum.

___ kg*m^2

(b) Calculate the torque about the fulcrum, using CCW as positive.

___ N*m

(c) Calculate the angular acceleration of the contraption, using CCW as positive.

(d) Calculate the linear acceleration of the right end of the rod, using up as positive.

___ m/s^2

Update:

A hollow sphere is attached to the end of a uniform rod. The sphere has a radius of 0.68 m and a mass of 0.48 kg. The rod has a length of 1.22 m and a mass of 0.56 kg. The rod is placed on a fulcrum (pivot) at X=0.44 m from the left end of the rod.

Update 2:

Could someone show me how to do this? Only variables would be fine. Relevance
• a) Calculate the moment of inertia J (click for graphical table) of the contraption around the fulcrum in kg*m^2

Jℓ = 0.44^2/3*0.56*0.44/1.22+0.48*(0.44+0.68)^2 = 0.615 kg*m^2

Jr = (1.22-0.44)^2/3*(1.22-0.44)/1.22*0.56 = 0.073 kg*m^2

J = Jℓ+Jr = 0.615+0.073 = 0.688 kg*m^2

(b) Calculate the torque about the fulcrum, using CCW as positive in N*m

Tℓ = (0.48*(0.68+0.44)+0.56*0.44/1.22*0.44/2)*9.806 = 5.707 N*m CCW

Tr = (0.56*(1.22-0.44)/1.22*(1.22-0.44)/2)*9.806 = 1.369 N*m CW

T = Tℓ-Tr = 5.707-1.369 = 4.338 N*m CCW

(c) Calculate the angular acceleration α of the contraption, using CCW as positive in rad/s^2

α = T/J = 4.338/0.688 = 6.308 rad/sec^2

(d) Calculate the linear acceleration a of the right end of the rod, using up as positive in m/s^2

a = α*(1.22-0.44) = 6.308*0.78 = 4.92 m/sec^2

• This isn't a question and this site is for asking questions, so you've been reported. Bye now! thanks for the 2pts