Question

A) A man holds a 185-N ball in his hand, with the forearm horizontal (see the figure). He can support the ball in this position because of the flexor muscle force , which is applied perpendicular to the forearm. The forearm weighs 18.2 N and has a center of gravity as indicated. Find (a) the magnitude of and the (b) magnitude and (c) direction (as a positive angle counterclockwise from horizontal) of the force applied by the upper arm bone to the forearm at the elbow joint.

B) The parallel axis theorem provides a useful way to calculate the moment of inertia I about an arbitrary axis. The theorem states that I = I_cm + Mh², where I_cm is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, M is the total mass of the object, and h is the perpendicular distance between the two axes. Use this theorem and information to determine the moment of inertia (kg·m²) of a solid cylinder of mass M = 5.60 kg and radius R = 6.00 m relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

C) A platform is rotating at an angular speed of 1.50 rad/s. A block is resting on this platform at a distance of 0.237 m from the axis. The coefficient of static friction between the block and the platform is 0.507. Without any external torque acting on the system, the block is moved toward the axis. Ignore the moment of inertia of the platform and determine the smallest distance from the axis at which the block can be relocated and still remain in place as the platform rotates.

D) A  rotating door is made from four rectangular glass panes, as shown in the figure. The mass of each pane is 65.0 kg. A person pushes on the outer edge of one pane with a force of F = 133 N that is directed perpendicular to the pane. Determine the magnitude of the door's angular acceleration.

Answer 1

In equilibrium net torque about the joint = 0

M*x - Wball*L - Warm*l = 0

M*0.051 - (185*0.33) - (18.2*(0.051+0.089)) = 0

M = 1247 N

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along vertical

Fnet = 0

Fv + M - Wball - Warm = 0

Fv + 1247 - 185 - 18.2 = 0

Fv = -1043.8 N

along horizantal

Fnet = 0

Fx = 0

magnitude of force at joint = F =sqrt(Fx^2+Fy^2) = 1043.8 N

direction = downwards

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B)

I = Icm + M*h^2

Icm = (1/2)*M*R^2

h = R

I = (1/2)*M*R^2 + M*R^2 = (3/2)*M*R^2

I = (3/2)*5.6*6^2 = 302.4 kg m^2

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(c)

momentum is conserved

r1^2*w1 = r2^2*w2

w2 = (r1/r2)^2*w1

us*m*g = m*r2*w2^2

us*g = r2*w1^2*((r1/r2)^4

us*g = w1^2*r1^4/r2^3

0.507*9.8 = 1.5*0.237^4/r2^3

r2 = 0.0984 m <<<==ANSWER