Finding acceleration of cylinder

[TSny]

For tangential direction: $mg\sin(30^o)-f_2=ma\cos(30^o)$
For normal direction: $N-mg\cos(30^o)=ma\sin(30^o)$

Looks good now? :)

Looks right. Also, Chet's method using horizontal and vertical components of force from the dog looks nice.

 

[Saitama]

Looks right. Also, Chet's method using horizontal and vertical components of force from the dog looks nice.

I had the following equation:
\sum F_x=f_1+f_2\cos(30^o)-N\sin(30^o)=4ma

$\because f_2=mg\sin(30^o)-ma\cos(30^o)$ and $N=ma\sin(30^o)+mg\cos(30^o)$
f_1+mg\sin(30^o)\cos(30^o)-ma\cos^2(30^o)-ma\sin^2(30^o)-mg\cos(30^o)\sin(30^o)=4ma
\Rightarrow f_1=5ma

Taking torque about CM of cylinder
\tau=f_1R-f_2R=I\alpha
(taking anticlockwise as the positive direction)
\Rightarrow R(5ma-mg\sin(30^o)-ma\cos(30^o))=2mR^2a/R
Solving this doesn't give me the right answer.

 

[TSny]

When you substituted for $f_2$ in the torque equation, make sure you didn't make a mistake with a sign.

If you take counterclockwise as positive for rotation, then does $\alpha = +a/R$ or does $\alpha = -a/R$?

 

[Saitama]

When you substituted for $f_2$ in the torque equation, make sure you didn't make a mistake with a sign.

If you take counterclockwise as positive for rotation, then does $\alpha = +a/R$ or does $\alpha = -a/R$?

Ah yes, thanks TSny, it now gives the right answer! Thank you for your time.

 

[TSny]

Ah yes, thanks TSny, it now gives the right answer! Thank you for your time.

You are welcome. Good work.