Rigid Body Plane Motion: Consider the following system where a circular object r

Question

Rigid Body Plane Motion: Consider the following system where a circular object rolls down an incline. If the object has a radius of 0.3 m and a mass of 25 kg, determine both the acceleration of the object's center of gravity and its angular acceleration if (A) the object is a sphere, (B) the object is a cylinder of length 0.3 m, and (C) the object is a thin disk. The slope's incline is at 30 degree above the horizontal, the coefficient of static friction between the object and the slope is mu_k = 0.20 and there is no slip between the object and the slope.

Explanation / Answer

here,

radius , r = 0.3 m

mass , m = 25 kg

theta = 30 degree

A)

for a sphere

the accelration , a = net force /effective mass

a = m * g * sin(theta) /( m+ I/r^2)

a = m * g * sin(theta) /( m + 0.4 * m * r^2/r^2)

a = 9.8 * sin(30) /1.4

a = 3.5 m/s^2

the accelration of centre of gravity is 3.5 m/s^2

the angular accelration , alpha = a /r

alpha = 3.5 /0.3 = 11.7 rad/s^2

B)

for a cyclinder

the accelration , a = net force /effective mass

a = m * g * sin(theta) /( m+ I/r^2)

a = m * g * sin(theta) /( m + 0.5 * m * r^2/r^2)

a = 9.8 * sin(30) /1.5

a = 3.27 m/s^2

the accelration of centre of gravity is 3.27 m/s^2

the angular accelration , alpha = a /r

alpha = 3.27 /0.3 = 10.89 rad/s^2

C)

for a thin disk

the accelration , a = net force /effective mass

a = m * g * sin(theta) /( m+ I/r^2)

a = m * g * sin(theta) /( m + 0.5 * m * r^2/r^2)

a = 9.8 * sin(30) /1.5

a = 3.27 m/s^2

the accelration of centre of gravity is 3.27 m/s^2

the angular accelration , alpha = a /r

alpha = 3.27 /0.3 = 10.89 rad/s^2

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.