Your idea to treat your prior information of 272 successes in 400 attempts does have fairly solid Bayesian justification.

The problem you are dealing with, as you recognized, is that of estimating a success probability $\theta$ of a Bernoulli experiment. The Beta distribution is the corresponding "conjugate prior". Such conjugate priors enjoy the "fictitious sample interpretation":

The Beta prior is
$$ 
\pi(\theta)=\frac{\Gamma(\alpha_0+\beta_0)}{\Gamma(\alpha_0)\Gamma(\beta_0)}\theta^{\alpha_0-1}(1-\theta)^{\beta_0-1}
$$
This can be interpreted as the information contained in a sample of size $\underline{n}=\alpha_0+\beta_0-2$ (loosely so, as $\underline{n}$ need not be integer of course) with $\alpha_0-1$ successes:
$$ 
\pi(\theta)=\frac{\Gamma(\alpha_0+\beta_0)}{\Gamma(\alpha_0)\Gamma(\beta_0)}\theta^{\alpha_0-1}(1-\theta)^{\underline{n}-(\alpha_0-1)}
$$
Hence, if you take $\alpha_0+\beta_0-2=400$ and $\alpha_0-1=272$, this corresponds to prior parameters $\alpha_0=273$ and $\beta_0=129$. "Halving" the sample would lead to prior parameters $\alpha_0=137$ and $\beta_0=65$. Now, recall that the prior mean and prior variance of the beta distribution are given by
The mean and the variance of the beta distribution are
$$
\mu=\frac{\alpha}{\alpha+\beta}\qquad\text{and}\qquad\sigma^2=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}
$$
Halving the sample keeps the prior mean (almost) where it is:

    alpha01 <- 273
    beta01 <- 129
    (mean01 <- alpha01/(alpha01+beta01))
    
    alpha02 <- 137
    beta02 <- 65
    (mean02 <- alpha02/(alpha02+beta02))

but increases the prior *variance* from 

    (priorvariance01 <- (alpha01*beta01)/((alpha01+beta01)^2*(alpha01+beta01+1)))
    [1] 0.0005407484

to

    (priorvariance02 <- (alpha02*beta02)/((alpha02+beta02)^2*(alpha02+beta02+1)))
    [1] 0.001075066

as desired.