Question

a heavy plank is moving with horizontal acceleration a=2 in S I units towards left.A hollow cylinder of mass m=3kg and radius 25cm is on upper surface of plank and moves without slipping find it angular acceleration [about axis in rad per second square ] of hollow cylinder?

Answer 1

A heavy plank is moving with horizontal acceleration $a = 2$ in SI units towards left. A hollow cylinder of mass $m = 3$ kg and radius $25$ cm is on upper surface of plank and moves without slipping. Find it angular acceleration [about axis in rad per second square] of hollow cylinder?

Solution

Descriptions:

$M$ mass of plank

$m$ mass of cylinder

$a$ acceleration of plank relative to earth

$a_{c}$ acceleration of cylinder relative to earth

$a_{c}^{\prime}$ acceleration of cylinder relative to plank

$F_{f}$ force of friction between plank and cylinder

$\varepsilon$ angular acceleration of cylinder

$J$ moment of inertia of cylinder

$\tau$ torque of cylinder

$$M a + m a _ {c} = F$$ $$M a = F - F _ {f}$$

From (1) and (2):

$$F _ {f} = m a _ {c}$$

The acceleration of cylinder relative to plank:

$$a _ {c} ^ {\prime} = \varepsilon R$$

The acceleration of cylinder relative to earth:

$$a _ {c} = a - a _ {c} ^ {\prime}$$

We can use known definitions:

$$(J \varepsilon = \tau ; \quad \tau = R F _ {f}; \quad J = m R ^ {2}) \Rightarrow (m R ^ {2} \varepsilon = R F _ {f}) \Rightarrow (\varepsilon = \frac {F _ {f}}{m R}) \qquad (6)$$

From (3) and (6):

$$a _ {c} = \frac {F _ {f}}{m}; \quad \varepsilon R = \frac {F _ {f}}{m} \qquad (7)$$

Substitute (4) and (7) into (5):

$$\left(\frac {F _ {f}}{m} = a - \frac {F _ {f}}{m}\right) \Rightarrow \left(F _ {f} = \frac {m a}{2}\right)$$

Finally after substitution of (8) in (6):

$$\varepsilon = \frac {a}{2 R} = 4 r a d / s ^ {2}$$

Answer: 4 rad/s²