Torque and acceleration. Show your work and explain your reasoning. A block with

Question

Torque and acceleration. Show your work and explain your reasoning. A block with mass m = 5.00 kg slides down a frictionless surface inclined 36.9 degrees to the. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has a mass 250 kg and moment of inertia 0.62 kg m^2 with respect to the axis of rotation. The string without slipping on the flywheel at a perpendicular distance of 0.200 m from axis. The free-body diagram for the wheel is shown in Figure (b). Explain why the angle shown for the normal force is correct in this situation. What is the acceleration of the block down the ramp in m/s^2 to the decimal places? What is the tension in the string in N to two decimal places?

Explanation / Answer

a. the normal force being a reaction force is always perpendicular to the point of contact between the object and the surface.

b.

Given:

mass of the block = m=5 kg

angle of inclination = 36.9 degree

so component of weight of block along the incline = mgsin36.9

flywheel moment of inertia = I = 0.62 kg. m^2

radius of pulley = R = 0.2 m

If a is the acceleration of the block down the ramp, the equations of motion can be written as below:

mg sin 36.9 - T = ma eqn 1

T X R = I x a/R   

so, T = I x a/R^2 eqn 2

substituting T value in eqn 1

mgsin 36.9 - I x a/R^2 = ma

or, mgsin 36.9 = ma + I x a/R^2

so, 5 x 9.8 x sin 36.9 = 5 x a + (0.62/0.2^2)a

so, 29.42 =20.5 a

so, a= acceleration of the block down the ramp = 29.42/20.5 = 1.44 m/s^2

c. Tension in the string= T = = I x a/R^2 = 0.62 x 1.44/(0.2)^2 = 22.32 N

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---